Lecture 12: Public-Key Cryptography and the RSA
Algorithm
Lecture Notes on “Computer and Network Security”
by Avi Kak ([email protected])
February 16, 2017
3:12pm
c
2017
Avinash Kak, Purdue University
Goals:
• To review public-key cryptography
• To demonstrate that confidentiality and sender-authentication can be
achieved simultaneously with public-key cryptography
• To review the RSA algorithm for public-key cryptography
• To present the proof of the RSA algorithm
• To go over the computational issues related to RSA
• To discuss the vulnerabilities of RSA
• Perl and Python implementations for generating primes and for
factorizing medium to large sized numbers
CONTENTS
Section Title
Page
12.1
Public-Key Cryptography
3
12.2
The Rivest-Shamir-Adleman (RSA) Algorithm for
Public-Key Cryptography — The Basic Idea
8
12.2.1
The RSA Algorithm — Putting to Use the Basic Idea
12
12.2.2
How to Choose the Modulus for the RSA Algorithm
14
12.2.3
Proof of the RSA Algorithm
17
12.3
Computational Steps for Key Generation in RSA
21
12.3.1
Computational Steps for Selecting the Primes p and q
22
12.3.2
Choosing a Value for the Public Exponent e
24
12.3.3
Calculating the Private Exponent d
27
12.4
A Toy Example That Illustrates How to Set n, e, and d
for a Block Cipher Application of RSA
28
12.5
Modular Exponentiation for Encryption and Decryption
34
12.5.1
An Algorithm for Modular Exponentiation
38
12.6
The Security of RSA — Vulnerabilities Caused by Lack
of Forward Secrecy
42
12.7
The Security of RSA — Chosen Ciphertext Attacks
45
12.8
The Security of RSA — Vulnerabilities Caused by LowEntropy Random Numbers
51
12.9
The Security of RSA — The Mathematical Attack
55
12.10
Factorization of Large Numbers: The Old RSA
Factoring Challenge
75
12.10.1
The Old RSA Factoring Challenge: Numbers Not Yet Factored
79
12.11
The RSA Algorithm: Some Operational Details
81
12.12
RSA: In Summary ....
92
12.13
Homework Problems
94
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12.1: PUBLIC-KEY CRYPTOGRAPHY
• Public-key cryptography is also known as asymmetric-key cryptography, to distinguish it from the symmetric-key cryptography
we have studied thus far.
• Encryption and decryption are carried out using two different
keys. The two keys in such a key pair are referred to as the
public key and the private key.
• With public key cryptography, all parties interested in secure
communications publish their public keys. [As to how that is done depends
on the protocol. In the SSH protocol, each server makes available through its port 22 the public key
it has stored for your login id on the server.
(See Section 12.10 for how an SSHD server acquires the
public key that the server would associate with your login ID so that you can make a password-free
connection with the server. In the context of the security made possible by the SSH protocol, the
public key held by a server is commonly referred to as the server’s host key.
)
When a client, such as
your laptop, wants to make a connection with an SSHD server, it sends a connection request to port 22
of the server machine and the server makes its host key available automatically. On the other hand, in
the SSL/TLS protocol, an HTTPS web server makes its public key available through a certificate of the
] As we will see, this solves one of the most
vexing problems associated with symmetric-key cryptography —
the problem of key distribution.
sort you’ll see in the next lecture.
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Lecture 12
• Party A, if wanting to communicate confidentially with party
B, can encrypt a message using B’s publicly available key. Such a
communication would only be decipherable by B as only B would
have access to the corresponding private key. This is illustrated
by the top communication link in Figure 1.
• Party A, if wanting to send an authenticated message to
party B, would encrypt the message with A’s own private key.
Since this message would only be decipherable with A’s public key, that would establish the authenticity of the message —
meaning that A was indeed the source of the message. This is
illustrated by the middle communication link in Figure 1.
• The communication link at the bottom of Figure 1 shows how
public-key encryption can be used to provide both confidentiality and authentication at the same time. Note again
that confidentiality means that we want to protect a message
from eavesdroppers and authentication means that the recipient needs a guarantee as to the identity of the sender.
• In Figure 1, A’s public and private keys are designated P UA and
P RA. B’s public and private keys are designated P UB and P RB .
• As shown at the bottom of Figure 1, let’s say that A wants to send
a message M to B with both authentication and confidentiality.
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Lecture 12
Party A wants to send a message to Party B
When only confidentiality is needed:
Party B
A’s private key
A’s public key
PRA
PUA
B’s public key
PUB
Encrypt with PU B
B’s private key
PRB
Decrypt with PR
B
Message
Message
Party A
When only authentication is needed:
Party B
A’s private key
A’s public key
PR A
PUA
B’s public key
PUB
Encrypt with PR A
B’s private key
PR B
Decrypt with PU
A
Message
Message
Party A
When both confidentiality and authentication are needed:
Encrypt
with PRA
Party B
A’s private key
A’s public key
PRA
PUA
B’s public key
PUB
B’s private key
PRB
Decrypt
with PRB
Encrypt
with PU
B
Decrypt
with PUA
Message
Message
Party A
Figure 1: This figure shows how public-key cryptography
can be used for confidentiality, for digital signatures, and
for both. (This figure is from Lecture 12 of “Computer and Network Security” by Avi Kak.)
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Lecture 12
The processing steps undertaken by A to convert M into its
encrypted form C that can be placed on the wire are:
C
=
E (P UB , E (P RA, M ))
where E() stands for encryption. The processing steps undertaken by B to recover M from C are
M
=
D (P UA, D (P RB , C))
where D() stands for decryption.
• The sender A encrypting his/her message with its own private key
P RA provides authentication. This step constitutes A putting
his/her digital signature on the message. Instead of applying the
private key to the entire message, a sender may also “sign” a message by applying
his/her private key to just a small block of data that is derived from the message to
be sent.
[DID YOU KNOW that you are required to digitally sign the software for your app before you
can market it through the official Android application store Google Play? And did you know that Apple’s App
]
Store has the same requirement?
• The sender A further encrypting his/her message with the
receiver’s public key P UB provides confidentiality.
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Lecture 12
• Of course, the price paid for achieving confidentiality and authentication at the same time is that now the message must be
processed four times in all for encryption/decryption. The message goes through two encryptions at the sender’s place and two
decryptions at the receiver’s place. Each of these four steps involves separately the computationally complex public-key
algorithm.
• IMPORTANT: Note that public-key cryptography does not
make obsolete the more traditional symmetric-key cryptography.
Because of the greater computational overhead associated with
public-key crypto systems, symmetric-key systems continue to
be widely used for content encryption. However, public-key encryption has proved indispensable for key management, for distributing the keys needed for the more traditional symmetric key
encryption/decryption of the content, for digital signature applications, etc.
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12.2: THE RIVEST-SHAMIR-ADLEMAN
(RSA) ALGORITHM FOR PUBLIC-KEY
CRYPTOGRAPHY — THE BASIC IDEA
• The RSA algorithm — named after Ron Rivest, Adi Shamir, and
Leonard Adleman — is based on a property of positive integers
that we describe below.
• As a direct consequence of the Euler’s Theorem of
Section 11.4 of Lecture 11, we can state that when a
and n are relatively prime, in arithmetic operations
am mod n, the exponents behave modulo the totient
φ(n) of n. [See Section 11.3 of Lecture 11 for the definition of the totient of a number.] Euler’s
theorem says that aφ(n) ≡ 1 (mod n) when a and n are relatively
prime. Given an arbitrary exponent k, we may express it as
k = k1φ(n) + k2 for some values of k1 and k2. Euler’s theorem
implies: ak mod n = ak2 mod n.
• For example, consider arithmetic modulo 15. As explained in
Section 11.3 of Lecture 11, since 15 = 3 × 5, we have φ(15) =
2×4 = 8 for the totient of 15. You can easily verify the following:
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47 · 44 mod 15 = 4(7+4) mod 8 mod 15 = 43 mod 15 = 64 mod 15 = 4
(43)5 mod 15 = 4(3×5) mod 8 mod 15 = 47 mod 15 = 4
Note that in both cases the base of the exponent, 4, is coprime
to the modulus 15.
• It follows from Euler’s theorem that given two exponents e and
d such that one is the multiplicative inverse of the other modulo
φ(n), we have M e×d ≡ M e×d (mod φ(n)) ≡ M (mod n).
• The result shown above, which follows directly from Euler’s theorem, requires that M and n be coprime. However, as will be
shown in Section 12.2.3, when n is a product of two
primes p and q, this result applies to all M , 0 ≤ M < n.
In what follows, let’s now see how this property can be used for
message encryption and decryption.
• Considering arithmetic modulo n, let’s say that e is an integer
that is coprime to the totient φ(n) of n. Further, say that d is
the multiplicative inverse of e modulo φ(n). These definitions of
the various symbols are listed below for convenience:
n
=
a modulus f or modular arithmetic
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Lecture 12
φ(n)
=
the totient of n
e
=
an integer that is relatively prime to φ(n)
[T his guarantees that e will possess a
multiplicative inverse modulo φ(n)]
d
=
an integer that is the multiplicative
inverse of e modulo φ(n)
• Now suppose we are given an integer M , 0 ≤ M < n, that represents our message, then we can transform M into another integer
C that will represent our ciphertext by the following modulo exponentiation:
C
M e mod n
=
At this point, it may seem rather strange that we would want to
represent any arbitrary plaintext message by an integer. But, it
is really not that strange. Let’s say you want a block cipher that
encrypts 1024 bit blocks at a time. Every plaintext block can
now be thought of as an integer M of value 0 ≤ M ≤ 21024 − 1.
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• We can recover back M from C by the following modulo operation
M
C d mod n
=
since
(M e)d (mod n) = M ed
(mod φ(n))
11
≡
M
(mod n)
Computer and Network Security by Avi Kak
Lecture 12
12.2.1: The RSA Algorithm — Putting to Use the
Basic Idea
• The basic idea described in the previous subsection can be used
to create a confidential communication channel in the manner
described here.
• An individual A who wishes to receive messages confidentially
will use the pair of integers {e, n} as his/her public key. At the
same time, this individual can use the pair of integers {d, n} as
the private key. The definitions of n, e, and d are as in the
previous subsection.
• Another party B wishing to send a message M to A confidentially
will encrypt M using A’s public key {e, n} to create ciphertext
C. Subsequently, only A will be able to decrypt C using his/her
private key {d, n}.
• If the plaintext message M is too long, B may choose to use RSA
as a block cipher for encrypting the message meant for A. As
explained by our toy example in Section 12.4, when RSA is used
as a block cipher, the block size is likely to be half the number of
bits required to represent the modulus n. If the modulus required,
say, 1024 bits for its representation, message encryption would be
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Lecture 12
based on 512-bit blocks. [While, in principle, RSA can certainly be used as a
block cipher, in practice, on account of its excessive computational overhead, it is more
likely to be used just for server authentication and for exchanging a secret session key.
A session key generated with the help of RSA-based encryption can subsequently be
used for content encryption using symmetric-key cryptography based on, say, AES.]
• The important theoretical question here is as to what conditions
if any must be satisfied by the modulus n for this M → C → M
transformation to work?
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Lecture 12
12.2.2: How to Choose the Modulus for the RSA
Algorithm
• With the definitions of d and e as presented in Section 12.2, the
modulus n must be selected in such a manner that the following
is guaranteed:
e d
M )
≡
M ed
≡
M
(mod n)
We want this guarantee because C = M e mod m is the encrypted form of the message integer M and decryption is carried
out by C d mod n.
• It was shown by Rivest, Shamir, and Adleman that we have this
guarantee when n is a product of two prime numbers:
n
=
p×q
f or some prime p and prime q
(1)
• The above factorization is needed because the proof of the algorithm, presented in the next subsection, depends on the following
two properties of primes and coprimes:
1. If two integers p and q are coprimes (meaning, relatively prime
to each other), the following equivalence holds for any two
integers a and b:
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Lecture 12
{a ≡ b (mod p) and a ≡ b (mod q)}
⇔
{a ≡ b (mod pq)}
(2)
This equivalence follows from the fact a ≡ b (mod p) implies a − b = k1p for some integer k1. But since we also
have a ≡ b (mod q) implying a − b = k2q, it must be the
case that k1 = k3 × q for some k3. Therefore, we can write
a − b = k3 × p × q, which establishes the equivalence. (Note
that this argument breaks down if p and q have common factors other than 1.) [We will use this property in the next subsection to
arrive at Equation (11) from the partial results in Equations (9) and (10).]
2. In addition to needing p and q to be coprimes, we also want
p and q to be individually primes. It is only when p and
q are individually prime that we can decompose the totient of
n into the product of the totients of p and q. That is
φ(n)
=
φ(p) × φ(q)
=
(p − 1) × (q − 1)
(3)
See Section 11.3 of Lecture 11 for a proof of this. [We will use
this property to go from Equation (5) to Equation (6) in the next subsection.]
• So that the cipher cannot be broken by an exhaustive search for
the prime factors of the modulus n, it is important that both p
and q be very large primes. Finding the prime factors of
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Computer and Network Security by Avi Kak
Lecture 12
a large integer is computationally harder than determining its primality.
• We also need to ensure that n is not factorizable by one of the
modern integer factorization algorithms. More on that later in
these notes.
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Lecture 12
12.2.3: Proof of the RSA Algorithm
• We need to prove that when n is a product of two primes p and q,
then, in arithmetic modulo n, the exponents behave modulo the
totient of n. We will prove this assertion indirectly by establishing
that when an exponent d is chosen as a mod φ(n) multiplicative
inverse of another exponent e, then the following will always be
true M e×d ≡ M (modn).
• Using the definitions of d and e as presented in Section 12.2, since
the integer d is the multiplicative inverse of the integer e modulo
the totient φ(n), we obviously have
e×d ≡ 1
(mod φ(n))
(4)
This implies that there must exist an integer k so that
e×d − 1
≡
0 (mod φ(n))
=
k × φ(n)
(5)
• It must then obviously be the case that φ(n) is a divisor of the
expression e×d − 1. But since φ(n) = φ(p)×φ(q), the totients
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Lecture 12
φ(p) and φ(q) must also individually be divisors of e × d − 1.
That is
φ(p) | (e × d − 1)
and
φ(q) | (e × d − 1)
(6)
The notation ‘|’ to indicate that its left argument is a divisor of
the right argument was first introduced at the end of Section 5.1
in Lecture 5.
• Focusing on the first of these assertions, since φ(p) is a divisor of
e × d − 1, we can write
e×d − 1
=
k1φ(p)
=
k1(p − 1)
(7)
for some integer k1.
• Therefore, we can write for any integer M :
M e×d mod p = M e×d
− 1 + 1
mod p = M k1(p
− 1)
×M mod p
(8)
• Now we have two possibilities to consider: Since p is a prime, it
must be the case that either M and p are coprimes or that M is
a multiple of p.
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Lecture 12
– Let’s first consider the case when M and p are coprimes. By
Fermat’s Little Theorem (presented in Section 11.2 of Lecture
11), since p is a prime, we have
Mp
− 1
≡
1 (mod p)
Since this conclusion obviously extends to any power of the
left hand side, we can write
M k1(p
− 1)
≡
1 (mod p)
Substituting this result in Equation (8), we get
M e×d mod p
=
M mod p
(9)
– Now let’s consider the case when the integer M is a multiple
of the prime p. Now obviously, M mod p = 0. This will also
be true for M raised to any power. That is, M k mod p = 0
for any integer k. Therefore, Equation (9) will continue to be
true even in this case.
• From the second assertion in Equation (6), we can draw an identical conclusion regarding the other factor q of the modulus n:
M e×d mod q
19
=
M mod q
(10)
Computer and Network Security by Avi Kak
Lecture 12
• We established in Section 12.2.2 that, when p and q are coprimes,
for any integers a and b if we have a ≡ b (mod p) and a ≡ b
(mod q), then it must also be the case that a ≡ b (mod pq).
Applying this conclusion to the partial results shown in Equations
(9) and (10), we get
M e×d mod n
20
=
M mod n
(11)
Computer and Network Security by Avi Kak
Lecture 12
12.3: COMPUTATIONAL STEPS FOR
KEY GENERATION IN RSA
CRYPTOGRAPHY
• The computational steps for key generation are
1. Generate two different primes p and q
2. Calculate the modulus n = p × q
3. Calculate the totient φ(n) = (p − 1) × (q − 1)
4. Select for public exponent an integer e such that 1 < e < φ(n)
and gcd(φ(n), e) = 1
5. Calculate for the private exponent a value for d such that
d = e−1 mod φ(n)
6. P ublic Key
= [e, n]
7. P rivate Key = [d, n]
• The next three subsections elaborate on these computational
steps.
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Lecture 12
12.3.1: Computational Steps for Selecting the Primes
p and q in RSA Cryptography
• You first decide upon the size of the modulus integer n. Let’s say
that your implementation of RSA requires a modulus of size B
bits.
• To generate the prime integer p;
– Using a high-quality random number generator (See Lecture
10 on random number generation), you first generate a random
number of size B/2 bits.
– You set the lowest bit of the integer generated by the above
step; this ensures that the number will be odd.
– You also set the two highest bits of the integer; this ensures
that the highest bits of n will be set. (See Section 12.4 for an
explanation of why you need to set the first two bits.)
– Using the Miller-Rabin algorithm described in Lecture 11, you
now check to see if the resulting integer is prime. If not, you
increment the integer by 2 and check again. This becomes the
value of p.
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Lecture 12
• You do the same thing for selecting q. You start with a randomly
generated number of size B/2 bits, and so on.
• In the unlikely event that p = q, you throw away your random
number generator and acquire a new one.
• For greater security, instead of incrementing by 2 when the MillerRabin test fails, you generate a new random number.
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Lecture 12
12.3.2: Choosing a Value for the Public Exponent e
• Recall that encryption consists of raising the message integer M
to the power of the public exponent e modulo n. This step is
referred to as modular exponentiation.
• The mathematical requirement on e is that gcd(e, φ(n)) = 1,
since otherwise e will not have a multiplicative inverse mod φ(n).
Since n = p × q, this requirement is equivalent to the two
requirements gcd(e, φ(p)) = 1 and gcd(e, φ(q)) = 1. In other
words, we want gcd(e, p − 1) = 1 and gcd(e, q − 1) = 1.
• For computational ease, one typically chooses a value for e that is
prime, has as few bits as possible equal to 1 for fast multiplication,
and, at the same time, that is cryptographically secure in the
sense described in the next bullet. Typical values for e are 3, 17,
and 65537 (= 216 + 1). Each of these values has only two bits
set, which makes for fast modular exponentiation. But
don’t forget the basic requirement on e that it must be relatively
prime to p − 1 and q − 1 simultaneously. Whereas p is prime,
p − 1 definitely is not since it is even. The same goes for q − 1. So
even if you wanted to, you may not be able to use a small integer
like 3 for e.
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Lecture 12
• Small values for e, such as 3, are considered cryptographically
insecure. Let’s say a sender A sends the same message M to
three different receivers using their respective public keys that
have the same e = 3 but different values of n. Let these values
of n be denoted n1, n2, and n3. Let’s assume that an attacker
can intercept all three transmissions. The attacker will see three
ciphertext messages: C1 = M 3 mod n1, C2 = M 3 mod n2,
and C3 = M 3 mod n3. Assuming that n1, n2, and n3 are
relatively prime on a pairwise basis, the attacker can use the
Chinese Remainder Theorem (CRT) of Section 11.7 of Lecture
11 to reconstruct M 3 modulo N = n1 × n2 × n3. (This assumes that
M < n n n , which is bound to be true since M < n , M < n , and M < n .) Having
reconstructed M 3, all that the attacker has to do is to figure out
the cube-root of M 3 to recover M . Finding cube-roots of even
large integers is not that hard.
(The Homework Problems section includes a
programming assignment that focuses on this issue.)
3
1
2
3
1
2
3
• Having selected a value for e, it is best to double check that
we indeed have gcd(e, p − 1) = 1 and gcd(e, q − 1) = 1
(since we want e to be coprime to φ(n), meaning that we want
e to be coprime to p − 1 and q − 1 separately). If either p or q
is found to not meet these two conditions on relative primality
of φ(p) and φ(q) vis-a-vis e, you must discard the calculated p
and/or q and start over. (It is faster to build this test into the
selection algorithm for p and q.) When e is a prime and greater
then 2, a much faster way to satisfy the two conditions is to
ensure
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Computer and Network Security by Avi Kak
Lecture 12
p mod e
q mod e
6=
6
=
1
1
• To summarize the point made above, you give priority to
using a particular value for e – such as a value like 65537
that has only two bits set. Having made a choice for the encryption integer e, you now find the primes p and q that, besides
satisfying all other requirements on these two numbers, also satisfy the conditions that the chosen e would be coprime to the
totients φ(p) and φ(q).
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Lecture 12
12.3.3: Calculating the Private Exponent d
• Once we have settled on a value for the public exponent e, the
next step is to calculate the private exponent d from e and the
modulus n.
• Recall that d × e ≡ 1 (mod φ(n)). We can also write this as
d
=
e−1 mod φ(n)
Calculating ‘e−1 mod φ(n)’ is referred to as modular inversion.
• Since d is the multiplicative inverse of e modulo φ(n), we can use
the Extended Euclid’s Algorithm (see Section 5.6 of Lecture 5)
for calculating d. Recall that we know the value for φ(n) since
it is equal to (p − 1) × (q − 1).
• Note that the main source of security in RSA is keeping p and q secret and therefore also keeping φ(n) secret. It is important to realize that knowing either will reveal
the other. That is, if you know the factors p and q, you can
calculate φ(n) by multiplying p − 1 with q − 1. And if you know
φ(n) and n, you can calculate the factors p and q readily.
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Lecture 12
12.4: A TOY EXAMPLE THAT
ILLUSTRATES HOW TO SET n, e, d FOR A
BLOCK CIPHER APPLICATION OF RSA
• As alluded to briefly at the end of Section 12.2.1, you are unlikely
to use RSA as a block cipher for general content encryption. As
mentioned in Section 12.12, for the moduli needed in today’s
computing environments, the computational overhead associated
with RSA is much too high for it to be suitable for content encryption. Nevertheless, RSA (along with ECC to be presented in
Lecture 14) plays a critical role in practically all modern protocols
for establishing secure communication links between clients and
servers. These protocols depend on RSA (and ECC) for clients
and servers to authenticate each other — as you’ll see in Lecture
13. In addition, RSA may also be used for generating session
keys. Despite the fact that you are not likely to use RSA for
content encryption, it’s nonetheless educational to reflect on how
it could be used for that purpose in the form of a block cipher.
• For the sake of illustrating how you’d use RSA as a block cipher,
let’s try to design a 16-bit RSA cipher for block encryption of disk
files. A 16-bit RSA cipher means that our modulus will span 16
bits. [Again, in the context of RSA, an N-bit cipher means that the modulus is of
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Lecture 12
size N bits and NOT that the block size is N bits. This is contrary to not-so-uncommon
usage of the phrase “N-bit block cipher” meaning a cipher that encrypts N-bit blocks
at a time as a plaintext source is scanned for encryption.]
• With the modulus size set to 16 bits, we are faced with the important question of what to use for the size of bit blocks for
conversion into ciphertext as we scan a disk file. Since our message integer M must be smaller than the modulus n, obviously
our block size cannot equal the modulus size. This requires that
we use a smaller block size, say 8 bits, and use some sort of a
padding scheme to fill up the rest of the 8 bits. As it turns out,
padding is an extremely important part of RSA ciphers. In addition to the need for padding as explained here, padding is also
needed to make the cipher resistant to certain vulnerabilities that
are described in Section 12.7 of this lecture.
• In the rest of the discussion in this section, we will assume for our
toy example that our modulus will span 16 bits, but the block
size will be smaller than 16 bits, say, only 8 bits. We will further
assume that, as a disk file is scanned 8 bits at a time, each such
bit block is padded on the left with zeros to make it 16 bits wide.
We will refer to this padded bit block as our message integer M .
• So our first job is to find a modulus n whose size is 16 bits. Recall
that n must be a product of two primes p and q. Assuming
that we want these two primes to be roughly the same size, let’s
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Lecture 12
allocate 8 bits to p and 8 bits to q.
• So the issue now is how to find a prime suitable for our 8-bit
representation. Following the prescription given in Section 12.3.1,
we could fire up a random number generator, set its first two
bits and the last bit, and then test the resulting number for its
primality with the Miller-Rabin algorithm presented in Lecture
11. But we don’t need to go to all that trouble for our toy
example. Let’s use the simpler approach described below.
• Let’s assume that we have an as yet imaginary 8-bit word for p
whose first two and the last bit are set. And assume that the same
is true for q. So both p and q have the following bit patterns:
bits of p
bits of q
11 − − − − − 1
11 − − − − − 1
:
:
where ’−’ denotes the bit that has yet to be determined. As you
can verify quickly from the three bits that are set, such an 8-bit
integer will have a minimum decimal value of 193. [Here is a reason
for why you need to manually set the first two bits: Assume for a moment that you
set only the first bit. Now it is theoretically possible for the smallest values for p and q
to be not much greater than 27 . So the product p × q could get to be as small as 214 ,
which obviously does not span the full 16 bit range desired for n. When you set the first
two bits, now the smallest values for p and q will be lower-bounded by 27 + 26 . So the
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Computer and Network Security by Avi Kak
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product p × q will be lower-bounded by 214 + 2 × 213 + 212 , which itself is lower-bounded
by 2 × 214 = 215 , which corresponds to the full 16-bit span. With regard to the setting
of the last bit of p and q, that is to ensure that p and q will be odd.]
• So the question reduces to whether there exist two primes (hopefully different) whose decimal values exceed 193 but are less than
255. If you carry out a Google search with a string like “first
1000 primes,” you will discover that there exist many candidates
for such primes. Let’s select the following two
p
q
=
=
197
211
which gives us for the modulus n = 197 × 211 = 41567. The bit
pattern for the chosen p, q, and modulus n are:
bits of p : 0Xc5
=
bits of q : 0Xd3
=
bits of n : 0Xa25f =
1100 0101
1101 0011
1010 0010 0101 1111
As you can see, for a 16-bit RSA cipher, we have a
modulus that requires 16 bits for its representation.
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Lecture 12
• Now let’s try to select appropriate values for e and d.
• For e we want an integer that is relatively prime to the totient
φ(n) = 196 × 210 = 41160. Such an e will also be relatively
prime to 196 and 210, the totients of p and q respectively. Since
it is preferable to select a small integer for e, we could try e = 3.
But that does not work since 3 is not relatively prime to 210. The
value e = 5 does not work for the same reason. Let’s try e = 17
because it is a small number and because it has only two bits
set.
• With e set to 17, we must now choose d as the multiplicative
inverse of e modulo 41160. Using the Bezout’s identity based
calculations described in Section 5.6 of Lecture 5, we write
gcd(17, 41160)
= gcd(41160, 17)
= gcd(17, 3)
= gcd(3,2)
= gcd(2,1)
|
|
|
|
|
|
|
|
|
|
|
residue 17 =
residue 3 =
residue 2 =
=
=
residue 1 =
=
=
=
0 x 41160 + 1 x 17
1 x 41160 - 2421 x 17
-5 x 3
+ 1 x 17
-5x(1 x 41160 - 2421 x 17) + 1 x 17
12106 x 17 - 5 x 41160
1x3 - 1 x 2
1x(41160 - 2421x17)
- 1x(12106x17 -5x41160)
6 x 41160
- 14527 x 17
6 x 41160
+ 26633 x 17
where the last equality for the residue 1 uses the fact that the
additive inverse of 14527 modulo 41160 is 26633. [If you don’t like
working out the multiplicative inverse by hand as shown above, you can use the Python
script FindMI.py presented in Section 5.7 of Lecture 5. Another option would be to
use the multiplicative inverse() method of the BitVector class.]
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Computer and Network Security by Avi Kak
Lecture 12
• The Bezout’s identity shown above tells us that the multiplicative
inverse of 17 modulo 41160 is 26633. You can verify this fact by
showing 17 × 26633 mod 41160 = 1 on your calculator.
• Our 16-bit block cipher based on RSA therefore has the following
numbers for n, e, and d:
n
e
d
=
=
=
41567
17
26633
Of course, as you would expect, this block cipher would have no
security since it would take no time at all for an adversary to
factorize n into its components p and q.
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Lecture 12
12.5: MODULAR EXPONENTIATION
FOR ENCRYPTION AND DECRYPTION
• As mentioned already, for encryption, the message integer M
is raised to the power e modulo n. That gives us the ciphertext
integer C. Decryption consists of raising C to the power d modulo
n.
• The exponentiation operation for encryption can be carried out
efficiently by simply choosing an appropriate e. (Note that the
only condition on e is that it be coprime to φ(n).) As mentioned
previously, typical choices for e are 3, 17, 35, 65537, etc. All
these integers have only a small number of bits set.
• Modular exponentiation for decryption, meaning the calculation
of C d mod n, is an entirely different matter since we are not
free to choose d. The value of d is determined completely by e
and n. Typically, d is of the same order as the modulus n.
• Computation of C d mod n can be speeded up by using the
Chinese Remainder Theorem (CRT) (see Section 11.7 of Lecture 11 for
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Computer and Network Security by Avi Kak
Lecture 12
CRT).
Since the party doing the decryption knows the prime factors p and q of the modulus n, we can first carry out the easier
exponentiations:
Vp
Vq
C d mod p
C d mod q
=
=
• To apply CRT as explained in Section 11.7 of Lecture 11, we must
also calculate the quantities
Xp
Xq
=
=
q × (q −1 mod p)
p × (p−1 mod q)
Applying CRT, we get
C d mod n
=
(VpXp + Vq Xq ) mod n
• Further speedup can be obtained by using Fermat’s Little Theorem (presented in Section 11.2 of Lecture 11) that says that if a
and p are coprimes then ap−1 mod p = 1.
• To see how Fermat’s Little Theorem (FLT) can be used to speed
up the calculation of Vp and Vq : Vp requires C d mod p. Since p
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Computer and Network Security by Avi Kak
Lecture 12
is prime, obviously C and p will be coprimes. We can therefore
write
Vp = C d mod p = C u×(p−1)
+ v
mod p = C v mod p
for some u and v. Since v < d, it’ll be faster to compute
C v mod p than C d mod p.
• When you use FLT in conjunction with CRT, you can calculate
C d (mod n) in roughly quarter of the time it takes otherwise. [First
note, as stated earlier in Section 12.3.1, both p and q are of the order of n/2 where n is the modulus. Since
Vp = C d (mod p) = C d
mod(p−1)
(mod p), and since d is of the order of n and d mod(p − 1) of the order
of p (which itself is of the order of n/2), it should take no more than half the number of multiplications to
calculate Vp compared to the number of multiplications needed for calculating C d (mod n) directly. The same
would be true for calculating Vq . As a result, the total number of multiplications required for both Vp and
Vq would be the same as in the direct calculation of C d (mod n). Note, however, the intermediate results in
the modular exponentiation needed for Vp would never exceed p (and the same would never exceed q for Vq ).
Since integer multiplication takes time that is proportional to the square of the size of the bit fields involved,
each multiplication involved in the calculation of Vp and Vq would take only one-quarter of the time it takes
]
for each multiplication in computing C d (mod n) directly.
• While the speedup achieved with CRT is impressive
indeed, it comes at a cost: It makes the calculation
of C d (mod n) vulnerable to different types of Side
Channel Attacks, such as the Fault Injection Attack
and the Timing Attack. In the Fault Injection attack, for example, you can get a processor to reveal
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Computer and Network Security by Avi Kak
Lecture 12
the values of the prime factors p and q just by deliberately causing the processor to miscalculate the
value of either Vp or Vq (but not both). See Lecture
32 on “Security Vulnerabilities of Mobile Devices”
for further information regarding these attacks.
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Lecture 12
12.5.1: An Algorithm for Modular Exponentiation
• After we have simplified the problem of modular exponentiation
considerably by using CRT and Fermat’s Little Theorem as discussed in the previous subsection, we are still left with having to
calculate:
AB mod n
for some integers A, B, and for some modulus n.
• What is interesting is that even for small values for A and B,
the value of AB can be enormous. Even when A and B consist
of only a couple of digits, as in 711, the result can still be a very
large number. For example, 711 equals 1, 977, 326, 743, a number
with 10 decimal digits. Now just imagine what would happen if,
as would be the case in cryptography, A has 256 binary digits
(that is 77 decimal digits) and B has 65537 binary digits. Even
when B has only 2 digits (say, B = 17), when A has 77 decimal
digits, AB will have 1304 decimal digits.
• The calculation of AB can be speeded up by realizing that if B
can be expressed as a sum of smaller parts, then the result is
a product of smaller exponentiations. We can use the following
binary representation for the exponent B:
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Computer and Network Security by Avi Kak
Lecture 12
B
≡
bk bk−1bk−2 . . . b0
(binary)
where we are saying that it takes k bits to represent the exponent,
each bit being represented by bi , with bk as the highest bit and
b0 as the lowest bit. In terms of these bits, we can write the
following equality for B:
B
=
2i
X
bi 6=0
• Now the exponentiation AB may be expressed as
B
A
P
A
=
bi 6=0 2
i
=
Y
A2
i
bi 6=0
We could say that this form of AB roughly halves the difficulty
of computing AB because, assuming all the bits of B are set, the
largest value of 2i will be about half the largest value of B.
• We can achieve further simplification by bringing the rules of
modular arithmetic into the multiplications on the right:
AB mod n
=

 Y

bi 6=0
"
i
A2 mod n

#


mod n
Note that as we go from one bit position to the next higher bit
position, we square the previously computed power of A.
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Computer and Network Security by Avi Kak
Lecture 12
i
• The A2 terms in the above product are of the following form
0
2
1
3
A2 , A2 , A2 , A2 , . . .
As opposed to calculating each term from scratch, we can calculate each by squaring the previous value. We may express this
idea in the following manner:
A, A2previous, A2previous, A2previous, . . .
• Now we can write an algorithm for exponentiation that scans the
binary representation of the exponent B from the lowest bit to
the highest bit:
result = 1
while B > 0:
if B & 1:
result = ( result * A ) % n
B = B >> 1
A = ( A * A ) % n
return result
# check the lowest bit of B
# shift B by one bit to right
• To see the dramatic speedup you get with modular exponentiation, try the following terminal session with Python
[ece404.12.d]$ => script
Script started on Mon 20 Feb 2012 10:23:32 PM EST
[ece404.12.d]$ => python
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Computer and Network Security by Avi Kak
Lecture 12
>>>
>>> print pow(7, 9633196, 9633197)
117649
>>>
>>>
>>>
>>> print (7 ** 9633196) % 9633197
117649
>>>
where the call to pow(7, 9633196, 9633197) calculates
79633197−1 mod 9633197 through Python’s implementation of the
modular exponentiation algorithm presented in this section. This
call will return instantaneously with the answer shown above.
On the other hand, the second call that carries out the same
calculation, but without resorting to modular exponentiation,
may take several minutes, depending on the hardware in your
machine. [You are encouraged to make similar comparisons with numbers that are even larger
than those shown here. If you wish, you can record your terminal-interactive Python session with the
command script as I did for the session presented above. First invoke script and then invoke
python as shown above. Your interactive work will be saved in a file called typescript. You can exit
the Python session by entering Ctrl-d and then exit the recording of your terminal session by entering
Ctrl-d again.
•
]
An important point to note is that whereas the RSA algorithm
is made theoretically possible by the number property stated in
Section 12.2, the algorithm is made practically possible by the
fact that there exist fast and memory-efficient algorithms for
modular exponentiation.
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Computer and Network Security by Avi Kak
Lecture 12
12.6: THE SECURITY OF RSA —
VULNERABILITIES CAUSED BY LACK
OF FORWARD SECRECY
• A communication link possesses forward secrecy if the content
encryption keys used in a session cannot be inferred from a future compromise of one or both ends of the communication link.
Forward secrecy is also referred to as Perfect Forward Secrecy.
• To see why RSA lacks forward secrecy, imagine a patient attacker
who is recording the encrypted communications between a server
and client.
• As you will see in Lecture 13, in order to establish an encrypted
session with a server (which could be an e-commerce website like Amazon.com), a client
(which could be your laptop) downloads the server’s certificate to, first, authenticate the server and to, then, get hold the server’s RSA
public key for the purpose of creating a secret session key. [As you
will learn in Lecture 13, a client generates a pseudorandom number to serve as the session key. To transmit
this session key to the server, the client encrypts it with the server’s public key so that only the server would
be able to decrypt it with its RSA private key. The client sends the encrypted session key to the server and,
]
subsequently, the two sides engage in an encrypted conversation.
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Computer and Network Security by Avi Kak
Lecture 12
• The attacker, who has managed to install a packet sniffer in the
LAN to which the client is connected, patiently records all encrypted communications between the client and the server with
the expectation that someday he will be able to get hold of the
server’s private keys. Obviously, if that were to happen, the attacker would be able to decrypt the session key that was sent
encrypted by the client to the server. And, as you can imagine,
after the attacker has figured out the session key, the attacker will
be able to decipher all of the recorded communications between
the client and the server.
• The attacker gaining access to a server’s private keys is not as
far fetched a scenario as one might think. Private keys may be
leaked out anonymously by disloyal employees or through bugs
in software. The Heartbleed bug that was discovered on April
7, 2014 is just the latest example of how private keys may fall
prey to theft through bugs in software. [See Section 20.4.4 of Lecture
20 for further information on the Heartbeat Extension to the SSL/TLS protocol and
the Heartbleed bug.]
• We say that the basic RSA algorithm makes it possible to carry
out the exploit described above because it lacks forward secrecy.
Whether or not this vulnerability in a given server-client interaction is a serious matter depends on the nature of the communications between the two — especially on the lifetime of the
information exchanged between the two endpoints.
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Computer and Network Security by Avi Kak
Lecture 12
• The solution to this problem with RSA lies in somehow creating a secret session key without putting it
on the wire. Naturally, your first reaction to this thought
would be: “but that is impossible!!!.” You are likely to
add: “How can two sides share a secret without either mentioning it to the other?”
• However, as they say, never underestimate the power of human
ingenuity. In Lecture 13, we will talk about an incredibly beautiful algorithm, known as the Diffie-Hellman (DH) algorithm, that
makes it possible to create a session key without either party
transmitting the key to the other party.
• Consequently, DH provides Perfect Forward Secrecy. However,
as you will see in Lecture 13, DH does suffer from a shortcoming of its own: it is vulnerable to the man-in-the-middle attack.
By combining RSA with DH, what you get — denoted DHE-RSA — gives you perfect forward secrecy
through the use of DH for exchanging the session
keys and RSA for endpoint (say, server) authentication. DHE stands for “Diffie-Hellman Exchange.” Another
commonly used combination protocol for creating secret session
keys is ECDHE-RSA where ECDHE stands for Elliptic Curve
Diffie-Hellman Exchange. The subject of elliptic curves for cryptography is presented in Lecture 14.
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Computer and Network Security by Avi Kak
Lecture 12
12.7: THE SECURITY OF RSA —
CHOSEN CIPHERTEXT ATTACKS
• The basic RSA algorithm — that is, an encryption/decryption
scheme whose implementation does not go beyond the mathematics of RSA as described so far — would be much too vulnerable
to all kinds of attacks, simple and fancy. Regarding the simpler
vulnerabilities, consider this: If we were to use the RSA algorithm only as
it has been described so far, think of the following vulnerability: Let’s say your public
key uses the exponent 3 and that you are in the habit of sending very short messages to
your business partners. If a message M is short enough, the ciphertext integer C = M 3
will be smaller than the modulus. Your enemies will be able to recover the plaintext
integer M simply by taking the cube-root of C by using, say, the nth root algorithm.
Such attacks become unfeasible when message integers are padded, in the manner described in this section, so as to span the full length of the modulus. With appropriate
padding, when the message M is raised to the power of the public exponent (even a
small public exponent like 3), the result would exceed the modulus and C would now be
the remainder modulo the modulus. Since nth root algorithm do not exist for modular
arithmetic, the enemy would not be able to recover M even if it is just a short message.
• Regarding the “fancier” vulnerabilities that RSA would fall prey
to if it were to be implemented just in the form described so
far, in this section we consider what are known as the Chosen
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Computer and Network Security by Avi Kak
Lecture 12
Ciphertext Attacks (CCA) on the RSA cipher.
• My immediate goal in this section is to convey to the reader what
is meant by CCA. As to how RSA is made secure against CCA is
a story of what goes into the padding bytes that are prepended
to the data bytes in order to create a block of bytes that spans
the width of the modulus.
• So that you understand the basic notion of CCA, a good place
to start this section is to show how the data bytes are padded
in Version 1.5 of the PKCS#1 scheme for RSA. This scheme is
also more compactly referred to by the string “PKCS#1v1.5”.
[Going beyond the fundamental notions of RSA public-key cryptography presented in
this lecture, how exactly those notions should be used in practice is governed by the
different PKCS “schemes.” The acronym PKCS stands for “Public Key Cryptography
Standard.” It designates a set of standards from RSA Labs for public-key cryptography.]
Despite the fact that Version 1.5 was promulgated in 1993,
I believe it is still the most widely used RSA scheme today. [Note
that Versions 2.0 and higher of the PKCS#1 scheme are resistant to all known forms of
CCA attacks. By the way, you can download all of the different versions of the PKCS#1
standard from the
]
http://www.rsa.com/rsalabs/ web site.
• In PKCS#1v1.5, what is subject to encryption is a block of bytes,
called, naturally, an Encryption Block (EB), that is composed of
the following sequence of bytes:
46
Computer and Network Security by Avi Kak
00
||
BT
Lecture 12
||
PS
||
<---------- k bytes
00
||
D
---------->
k = size of modulus in bytes
where ‘||’ means simple concatenation, the numeric ‘00’ stands
for a byte whose value is 0, the notation ‘BT’ means a one-byte
integer that designates the type of EB, the notation ‘PS’ means
a pseudorandomly generated “Padding String”, and the symbol
‘D’ stands for the data bytes. The value of ‘BT’ is the integer 2
for encryption with RSA. [The values 0 and 1 for ‘BT’ are meant for RSA when it is used
for digital signatures.]
• The PKCS#1v1.5 standard mandates that the pseudorandomly
generated padding string PS contain at least 8 bytes for security
reasons. Therefore, in PKCS#1v1.5, the minimum value for k,
the size of the modulus, is 12 bytes. That would be accounted
for by one byte for ’00’, one for ’BT’, 8 for ’PS’, one for another
’00’, with one leftover for the data byte ’D’.
• With that brief introduction to how an encryption block is constructed in PKCS#1v1.5, let’s get back to the subject of CCA.
• Let’s say you use my public key (n, e) to encrypt a plaintext
message M into the ciphertext C. You send C to me, but on its
way to me, the ciphertext C is picked up by someone we’ll refer
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Computer and Network Security by Avi Kak
Lecture 12
to as the attacker. Through CCA, the attacker can figure out
what the plaintext message M is even without having to know
the decryption exponent d. The attacker’s exploit would consist
of the following steps:
– The attacker randomly chooses an integer s.
– The attacker constructs a new message — which hopefully would not
arouse my suspicion — by forming the product C ′ = se × C mod n.
– The attacker somehow lures me into decrypting C ′ . (I may cooperate
because C ′ looks innocuous to me.)
– Assume that, for whatever reason, I send back to the attacker M ′ =
C ′ d = (se × C)d mod n = se×d × C d mod n = s × M mod n.
– The attacker will now be able to recover the original message M by
M = M ′ × s−1 mod n, assuming that the multiplicative inverse of
s exists in Zn . Remember, the choice of s is under the attacker’s
control.
• The fact that RSA could be vulnerable to such attacks was first
discovered by George Davida in 1982.
• Another form of CCA was discovered by Daniel Bleichenbacher
in 1998. In this attack, the attacker uses a sequence of randomly
selected integers s to form a candidate sequence of ciphertexts
C ′ = se × C mod n. The attacker chooses the integers s one
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Computer and Network Security by Avi Kak
Lecture 12
at a time, forms the ciphertext C ′ , and sends it to an oracle just
to find out if C ′ is likely to have been produced by a message
whose first two bytes have the integer values of 0 and 2 — in
accordance with the format of the encryption block shown earlier in this section. Each positive return from the oracle allows
the attacker to enlarge the size of s and make an increasingly
narrower estimate for the value of the plaintext integer M that
corresponds to the original C. The iterations end when the estimated value for M is just one number. [In case you are wondering about
the “oracle” and as to what that would correspond to in practice, the goal here is merely to demonstrate
that the attacker can recover the message integer M even with very limited knowledge that consists of some
mechanism informing the attacker whether or not the chosen C ′ violates the structure of the encryption block
that is stipulated for PKCS#1v1.5. Whether or not such a mechanism exists today is not the point. Such a
mechanism could consist of the victim’s RSA engine simply returning an error report whenever it receives a
ciphertext that it believes was produced by a message that did not conform to the encryption block structure in
] Bleichenbacher’s attack is reported in the publication
“Chosen Ciphertext Attacks Against Protocols Based on the
RSA Encryption Standard PKCS#1,” that is available from
his home page.
PKCS#1v1.5.
• These days one makes a distinction between two different types
of chosen ciphertext attacks and these are referred to as CCA1
and CCA2. Under CCA1, the attacker can consult the decryption oracle mentioned above an arbitrary number of times, but
only until the attacker has acquired the ciphertext C through
eavesdropping or otherwise. And, under CCA2, the attacker can
continue to consult the oracle even after seeing C. For obvious
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Computer and Network Security by Avi Kak
Lecture 12
reasons, in either model, the attacker cannot query the oracle
with C itself. The CCA1 attack is also known as the passive
chosen ciphertext attack and CCA2 as the adaptive chosen ciphertext attack. The attack by Bleichenbacher is an example of
CCA2. The success of that attack implies that PKCS#1v1.5 is
not CCA2 secure.
• RSA is made resistant to CCA2 when the padding bytes are
set according to OAEP. OAEP stands for Optimal Asymmetric
Encryption Padding. Unlike what you saw for the PKCS#1v1.5
format for encryption blocks at the beginning of this section, there
is no structure in the encryption blocks under PKCS#1v2.x. The
padding now involves a mask generation function that depends on
a hash applied to a set of parameters. For further information, the
reader is referred to the RSA Labs publication “RSAES-OAEP
Encryption Scheme” and the references contained therein. This
publication can be download from the same web site as mentioned
at the beginning of this section.
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Lecture 12
12.8: THE SECURITY OF RSA —
VULNERABILITIES CAUSED BY
LOW-ENTROPY RANDOM NUMBERS
• Please review Section 10.8 of Lecture 10 to appreciate the significance of “Low Entropy” in the title of this section. [As explained there,
the entropy of a random number generator is at its highest if all numbers are equally likely to be produced
within the range of numbers that the output is designed for. For example, if a CSPRNG can produce 512-bit
random numbers with equal probability, its entropy is at its maximum and it equals 512 bits. However, should
the probabilities associated with the output random numbers be nonuniform, the entropy will be less than 512.
The greater the nonuniformity of this probability distribution, the smaller the entropy. The entropy is zero for
]
deterministic output.
• Consider the following mind-boggling fact: If an attacker can
get hold of a pair of RSA moduli, N1 and N2, that
share a factor, the attacker will be able to figure
out the other factor for both moduli with hardly any
work. Obviously, once the attacker has acquired both factors
of a modulus, the attacker can quickly calculate the private key
that goes with the public key associated with the modulus. This
exploit, if successfully carried out, immediately yields the private
keys that go with the public keys that contain the N1 and N2
moduli.
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Computer and Network Security by Avi Kak
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• To see why that is the case, let’s say that p is the common factor
of the two moduli N1 and N2. That makes p the GCD of N1
and N2 . Now let’s denote the other factor in N1 by q1 and in N2
by q2 . You already know from Lecture 5 that Euclid’s recursion
makes the calculation of the GCD of any two numbers extremely
fast. [Using Euclid’s algorithm, the GCD of two 1024-bit integers on a routine desktop can be computed
in just a few microseconds using the Gnu Multiple Precision (GMP) library. More theoretically speaking, the
]
Therefore,
the common factor p of two moduli — assuming they have a
common factor — can be calculated almost instantaneously with
ordinary hardware. And once you have p, the factors q1 and q2
are obtainable by simple integer division, which is also fast.
computational complexity of Euclid’s GCD algorithm is O(n2 ) for n bit numbers.
• You might ask: Is it really likely that an attacker would find a
pair of RSA moduli that share a common factor? The answer is:
It is very, very likely today. Read on for why.
• Modern port and vulnerability scanners of the sort I’ll present
in Lecture 23 can carry out a full SSL/TLS and SSH handshake
and fetch the certificates used by the TLS/SSL hosts (these are
typically HTTPS web servers) and the host keys used by the
SSHD servers at a fairly rapid rate.
• In a truly landmark investigation by Nadia Heninger, Zakir Durumeric, Eric Wustrow, and J. Alex Halderman that was presented at the 2012 USENIX Security Symposium, the authors re52
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Lecture 12
ported harvesting over 5 million TLS/SSL certificates and around
4 million RSA-based SSH host keys through scans that lasted no
more than a couple of days. As you can see, these authors were
able to harvest a very large number of RSA moduli in a rather
short time. Subsequently they set out to find the factors that
any of the moduli shared with any of the other moduli. [While the
GCD of a pair of numbers can be computed very fast on a run-of-the-mill machine, it
would still take a very long time to do pairwise computation for all the numbers in a
set that contains a few million numbers. For further speedup, Heninger et al. used a
method proposed by Daniel Bernstein. In this method, you start with calculating the
product of all the moduli, multiplying two moduli at a time in what’s called a product
tree, and then reduce the product with respect to the pairwise products of the squares
of the moduli in what’s known as the remainder tree. This approach, applied to over
11 million RSA moduli from the TLS/SSL and SSH datasets, yielded the p factors in
under 6 hours on a multicore PC class machine with 32 GB of RAM.]
The title of
the publication by Heninger et al. is “Mining your Ps and Qs:
Detection of Widspread Weak Keys in Network Devices”.
• In this manner, Heninger et al. were able to compute the private
keys for 0.50% of the TLS/SSL servers (the HTTPS web servers)
and for 0.03% of the SSH servers.
• The upshot of the investigation reported by Heninger et al. is that
your random number generator must have high enough entropy
so that each modulus is unique vis-a-vis the moduli used by any
other communication device any place on the face of this earth!
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Computer and Network Security by Avi Kak
Lecture 12
• While the prescription stated above is followed for the most part
by most computers of the sort we use everyday, that’s not necessarily the case for a large number of what are known as headless
communication devices in the internet. By headless devices we
mean routers, firewalls, sever management cards, etc. As observed by Heninger et al., a very large number of such headless
devices use software entropy sources for the random bytes they
need as candidates for the prime numbers and the most commonly used software entropy source is /dev/urandom that supplies
pseudorandom bytes through non-blocking reads.
• The problem with /dev/urandom arises at boot time when such a
software entropy source is least equipped to supply high-entropy
random bytes and this happens exactly when the network interface has a need to create keys for communicating with other
hosts.
• See Section 10.9.4 of Lecture 10 for further information on /dev/urandom
and its relationship to /dev/random.
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Lecture 12
12.9: THE SECURITY OF RSA — THE
MATHEMATICAL ATTACK
• Assuming that the security issues brought up in the previous
three sections are not relevant in a given application context, the
security of RSA depends critically on the fact that whereas it is
easy to multiply two large primes to construct a modulus, the
inverse operation of factoring the modulus into its prime factors
can be extremely difficult — difficult until you solve the integer
factorization problem for the sizes of the numbers involved. [Functions
that are easy to compute in one direction but that cannot be easily inverted without special information
] Trying to break RSA by developing an
integer factorization solution for the moduli involved is known as
a mathematical attack.
are known as trapdoor functions.
• That is, a mathematical attack on RSA consists of figuring out
the prime factors p and q of the modulus n. Obviously, knowing p
and q, the attacker will be able to figure out the private exponent
d for decryption.
• Another way of stating the same as above would be that the
attacker would try to figure out the totient φ(n) of the modulus n.
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Computer and Network Security by Avi Kak
Lecture 12
But as stated earlier, knowing φ(n) is equivalent to knowing the
factors p and q. If an attacker can somehow figure out φ(n), the
attacker will be able to set up the equation (p−1)(q−1) = φ(n),
that, along with the equation p × q = n, will allow the attacker
to determine the values for p and q.
• Because of their importance in public-key cryptography, a number that is a product of two (not necessarily distinct) primes is
known as a semiprime. Such numbers are also called biprimes,
pq-numbers, and 2-almost primes. Currently the largest
known semiprime is
(230,402,457 − 1)2
This number has over 18 million digits. This is the square of the
largest known prime number.
• Over the years, various mathematical techniques have been developed for solving the integer factorization problem involving
large numbers. A detailed presentation of integer factorization
is beyond the scope of this lecture. We will now briefly mention
some of the more prominent methods, the goal here being
merely to make the reader familiar with the existence
of the methods. For a full understanding of the mentioned
methods, the reader must look up other sources where the methods are discussed in much greater detail [Be aware that while the methods listed
below can factorize large numbers, for very large numbers of the sort used these days in RSA cryptography,
you have to custom design the algorithms for each attack. Customization generally consists of making various
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Computer and Network Security by Avi Kak
Lecture 12
conjectures about the modulo properties of the factors and using the conjectures to speed up the search for the
]:
factors.
Trial Division: This is the oldest technique. Works quite well
for removing primes from large integers of up to 12 digits (that
is, numbers smaller then 1012). As the name implies, you simply divide the number to be factorized by successively larger
integers. A variation is to form a product m = p1p2p3 . . . pr
of r primes and to then compute gcd(n, m) for finding the
largest prime factor in n. Here is a product of all primes
p ≤ 97:
2305567963945518424753102147331756070
Fermat’s Factorization Method: Is based on the notion that
every odd number n that has two non-trivial factors can be
expressed as a difference of two squares, n = (x2 − y 2 ). If
we can find such x and y, then the two factors of n are (x − y)
and (x + y). Searching for these factors boils down to solving
x2 ≡ y 2 (mod n). This is referred to as a congruence of
squares. That every odd n can be expressed as a difference
of two squares follows from the fact that if n = a × b, then
n = [(a + b)/2]2
− [(a − b)/2]2
Note that since n is assumed to be odd, both a and b are
odd, implying that a + b and a − b will both be even. In its
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Computer and Network Security by Avi Kak
Lecture 12
implementation, one tries various values of x hoping to find
one that yields a square for x2 − n. The search is begun with
√
with the integer x = ⌈ n⌉. Here is the pseudocode for this
approach
x = ceil( sqrt( n ) )
# assume n is odd
y_squared = x ** 2 - n
while y_squared is not a square
x = x + 1
y_squared = x ** 2 - n
# y_squared = y_squared + 2*x + 1
return x - sqrt( b_squared )
This method works fast if n has a factor close to its squareroot. In general, its complexity is O(n). Fermat’s method can
be speeded up by using trial division for candidate factors up
√
to n.
Sieve Based Methods: Sieve is a process of successive crossing out entries in a table of numbers according to a set of
rules so that only some remain as candidates for whatever one
is looking for. The oldest known sieve is the sieve of Eratosthenes for generating prime numbers. In order to find
all the prime integers up to a number, you first write down
the numbers successively (starting with the number 2) in an
array-like display. The sieve algorithm then starts by crossing
out all the numbers divisible by 2 (and adding 2 to the list of
primes). Next you cross out all the entries in the table that are
divisible by 3 and you add 3 to the list of primes, and so on.
Modern sieves that are used for fast factorization are known as
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Computer and Network Security by Avi Kak
Lecture 12
quadratic sieve, number field sieve, etc. The quadratic
sieve method is the fastest for integers under 110 decimal digits and considerably simpler than the number field sieve. Like
the principle underlying Fermat’s factorization method, the
quadratic sieve method tries to establish congruences modulo
n. In Fermat’s method, we search for a single number x so
that x2 mod n is a square. But such x’s are difficult to find.
With quadratic sieve, we compute x2 mod n for many x’s and
then find a subset of these whose product is a square.
Pollard-ρ Method: It is based on the following observations:
– Say d is a factor of n. Obviously, the yet unknown
d satisfies d|n. Now assume that we have two randomly
chosen numbers a and b so that a ≡ b (mod d). Obviously,
for such a and b, a − b ≡ 0 (mod n), implying a − b = kd
for some k, further implying that d must also be a divisor of
the difference a−b. That is, d|(a−b). Since, by assumption,
d|n, it must be the case that gcd(a−b, n) is a multiple of d.
We can now set d to the answer returned by gcd, assuming
that this answer is greater than 1. Once we find such a
factor of n, we can divide n by the factor and repeat the
algorithm on the resulting smaller integer.
– This suggests the following approach to finding a factor of
√
n: (1) Randomly choose two numbers a, b ≤ n; (2) Find
gcd(a − b, n); (3) If this gcd is equal to 1, go back to step
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Computer and Network Security by Avi Kak
Lecture 12
1 until the gcd calculation yields a number d greater than
1. This d must be a factor of n.
[A discerning reader might say
that since we know nothing about the factor d of n and since we are essentially shooting in
the dark when making guesses for a and b, why should we expect a performance any better
than making random guesses for the factors of n up to the square-root of n. That may well
be true in general, but the beauty of searching for the factors via the differences a − b is
that it generalizes to the main feature of the Pollard-ρ algorithm in which the sequence
of integers you choose for b grows twice as fast as the sequence of integers you
choose for a. It is this feature that makes for a much more efficient way to look for the
factors of n. This feature is implemented in lines (E10), (E11), and (E12) of the code shown
at the end of this section. As was demonstrated by Pollard, letting b grow twice as fast as a
in gcd(a − b, n) makes for fast detection of cycles, these being two different numbers a and b
that are congruent modulo some integer d < n.
]
– In the code shown at the end of this section, the simple procedure laid out above is called pollard rho simple(); its
implementation is shown in lines (D1) through (D15) of the
code. We start the calculation by choosing random numbers for a and b, and computing gcd(a − b, n). Assuming
that this gcd equals 1, we now generate another candidate
for b in the loop in lines (D9) through (D14). For each
new candidate generated for b, its difference must be computed from all the previously generated random numbers
and the gcd of the differences computed. In general, for the
k th random number selected for b, you have to carry out k
calculations of gcd.
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Computer and Network Security by Avi Kak
Lecture 12
– The above mentioned ever increasing number of gcd calculations for each iteration of the algorithm is avoided by
what is the heart of the Pollard-ρ algorithm. The candidate
numbers are generated pseudorandomly using a function f
that maps a set to itself through the equivalence of the remainders modulo n. Let’s express the sequence of numbers
generated through such a function by xi+1 = f (xi) mod n.
Again assuming the yet unknown factor d of n, suppose we discover a pair of indices i and j, i < j, for
this sequence such that xi ≡ xj (mod d), then obviously
f (xi ) ≡ f (xj ) (mod d). This implies that each element of
the sequence after j will be congruent to each corresponding element of the sequence after i modulo the unknown
d.
– So let’s say we can find two numbers in the sequence xi and
x2i that are congruent modulo the unknown factor d, then
by the logic already explained d|(xi − x2i). Since d|n, it
must be case that gcd(xi − x2i, n) must be a factor of n.
– The Pollard-ρ algorithm uses a function f () to generate
two sequence xi and yi , with the latter growing twice as
fast as the former — see lines (E10), (E11), and (E12) of
the code for an illustration of this idea. That is, at each
iteration, the first sequence corresponds to xi+1 ← f (xi)
and yi+1 ← f (f (yi )). This would cause each (xi, yi ) pair
to be the same as (xi, x2i). If we are in the cycle part of the
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Computer and Network Security by Avi Kak
Lecture 12
sequence, and if xi ≡ x2i (mod d), then we must have a
d = gcd((xi − yi ), n), d 6= 1 and we are done.
– The most commonly used function f (x) is the polynomial
f (x) = x2 + c mod n with the constant c not allowed
to take the values 0 and −2. The code shown in lines
(E4) through (E15) constitutes an implementation of this
polynomial.
– Some parts of the implementation of the overall integer factorization algorithm shown below should already be familiar to you. The calculation of gcd in lines in (B1) through
(B4) is from Section 5.4.5 of Lecture 5. The Miller-Rabin
based primality testing code in lines (C1) through (C22) is
from Section 11.5.5 of Lecture 11.
#!/usr/bin/env python
##
##
##
##
Factorize.py
Author: Avi Kak
Date: February 26, 2011
Modified: Febrary 25, 2012
##
##
Uncomment line (F9) and comment out line (F10) if you want to see the results
with the simpler form of the Pollard-Rho algorithm.
import random
import sys
def factorize(n):
prime_factors = []
factors = [n]
while len(factors) != 0:
p = factors.pop()
if test_integer_for_prime(p):
prime_factors.append(p)
continue
#(F1)
#(F2)
#(F3)
#(F4)
#(F5)
#(F6)
#(F7)
#(F8)
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Computer and Network Security by Avi Kak
#
Lecture 12
d = pollard_rho_simple(p)
d = pollard_rho_strong(p)
if d == p:
factors.append(d)
else:
factors.append(d)
factors.append(p//d)
return prime_factors
#(F9)
#(F10)
#(F11)
#(F12)
#(F13)
#(F14)
#(F15)
#(F16)
def test_integer_for_prime(p):
probes = [2,3,5,7,11,13,17]
for a in probes:
if a == p: return 1
if any([p % a == 0 for a in probes]): return 0
k, q = 0, p-1
while not q&1:
q >>= 1
k += 1
for a in probes:
a_raised_to_q = pow(a, q, p)
if a_raised_to_q == 1 or a_raised_to_q == p-1: continue
a_raised_to_jq = a_raised_to_q
primeflag = 0
for j in range(k-1):
a_raised_to_jq = pow(a_raised_to_jq, 2, p)
if a_raised_to_jq == p-1:
primeflag = 1
break
if not primeflag: return 0
probability_of_prime = 1 - 1.0/(4 ** len(probes))
return probability_of_prime
#(P1)
#(P2)
#(P3)
#(P4)
#(P5)
#(P6)
#(P7)
#(P8)
#(P9)
#(P10)
#(P11)
#(P12)
#(P13)
#(P14)
#(P15)
#(P16)
#(P17)
#(P18)
#(P19)
#(P20)
#(P21)
#(P22)
def pollard_rho_simple(p):
probes = [2,3,5,7,11,13,17]
for a in probes:
if p%a == 0: return a
d = 1
a = random.randint(2,p)
random_num = []
random_num.append( a )
while d==1:
b = random.randint(2,p)
for a in random_num[:]:
d = gcd( a-b, p )
if d > 1: break
random_num.append(b)
return d
#(Q1)
#(Q2)
#(Q3)
#(Q4)
#(Q5)
#(Q6)
#(Q7)
#(Q8)
#(Q9)
#(Q10)
#(Q11)
#(Q12)
#(Q13)
#(Q14)
#(Q15)
def pollard_rho_strong(p):
probes = [2,3,5,7,11,13,17]
for a in probes:
if p%a == 0: return a
d = 1
a = random.randint(2,p)
c = random.randint(2,p)
#(R1)
#(R2)
#(R3)
#(R4)
#(R5)
#(R6)
#(R7)
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Computer and Network Security by Avi Kak
Lecture 12
b = a
while d==1:
a = (a * a + c) % p
b = (b * b + c) % p
b = (b * b + c) % p
d = gcd( a-b, p)
if d > 1: break
return d
#(R8)
#(R9)
#(R10)
#(R11)
#(R12)
#(R13)
#(R14)
#(R15)
def gcd(a,b):
while b:
a, b = b, a%b
return a
#(S1)
#(S2)
#(S3)
#(D4)
if __name__ == ’__main__’:
#(A1)
if len( sys.argv ) != 2:
sys.exit( "Call syntax: Factorize.py number" )
p = int( sys.argv[1] )
factors = factorize(p)
print("\nFactors of %d:" % p)
for num in sorted(set(factors)):
print("%s %d ^ %d" % ("
", num, factors.count(num)))
#(A2)
#(A3)
#(A4)
#(A5)
#(A6)
#(A7)
#(A8)
– Let’s try the program on what is known as the sixth Fermat
number [The n Fermat number is given by 2 + 1. So the sixth Fermat number is 2 + 1.]:
2n
th
64
Factorize.py 18446744073709551617
The factors returned are:
274177 ^ 1
67280421310721 ^ 1
In the answer shown what comes after ^ is the power of the
factor in the number. You can check the correctness of the answer by entering the
number in the search window at the http://www.factordb.com web site. You will also
notice that you will get the same in only another blink of the
eye if you comment out line (F10) and uncomment line (F9),
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Computer and Network Security by Avi Kak
Lecture 12
which basically amounts to making a random guess for the
factors.
– That we get the same performance regardless of whether we
use the statement in line (F9) or the statement in line (F10)
happens because the number we asked Factorize.py to factorize above was easy. As we will mention in Section 12.9,
factorization becomes harder when a composite is a product
of two primes of roughly the same size. For that reason, a
tougher problem would be to factorize the known semiprime
10023859281455311421. Now, unless you are willing to wait
for a long time, you will have no choice but to use the statement in line (F10). Using the statement in line (F10), the
factors returned for this number are:
1308520867 ^ 1
7660450463 ^ 1
– For another example, when we call Factorize.py on the
number shown below, using the statement in line (F10) for
the Pollard-ρ algorithm
11579208923731619542357098500868790785326998466564056403
the factors returned are:
23 ^ 1
41 ^ 1
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Computer and Network Security by Avi Kak
Lecture 12
149 ^ 1
40076041 ^ 1
713526132967 ^ 1
9962712838657 ^ 1
289273479972424951 ^ 1
– Shown next is a Perl version of the script for factorization.
Since arbitrarily sized integers are not native to Perl, this
script can only handle integers that can be accommodated in
4 bytes that Perl uses for storing unsigned integers. [As mentioned
previously in Lecture 11, in Perl you must import the Math::BigInt package
for arbitrarily large integers. Later in this section I will show an implementation
of the Pollard-Rho factorization algorithm that is based on the Math::BigInt
representation of large integers.]
#!/usr/bin/env perl
##
##
##
Factorize.pl
Author: Avi Kak
Date: February 19, 2016
##
##
Uncomment line (F12) and comment out line (F13) if you want to see the results
with the simpler form of the Pollard-Rho algorithm.
use strict;
use warnings;
die "\nUsage:
$0 <integer> \n" unless @ARGV == 1;
my $p = shift @ARGV;
#(A1)
#(A2)
die "Your number is too large for factorization by this script. " .
"Instead, try the script ’FactorizeWithBigInt.pl’\n"
if $p > 0x7f_ff_ff_ff;
#(A3)
my @factors = @{factorize($p)};
my %how_many_of_each;
map {$how_many_of_each{$_}++} @factors;
print "\nFactors of $p:\n";
foreach my $factor (sort {$a <=> $b} keys %how_many_of_each) {
print "
$factor ^ $how_many_of_each{$factor}\n";
}
66
#(A4)
#(A5)
#(A6)
#(A7)
#(A8)
#(A9)
Computer and Network Security by Avi Kak
Lecture 12
sub factorize {
my $n = shift;
my @prime_factors = ();
my @factors;
push @factors, $n;
while (@factors > 0) {
my $p = pop @factors;
if (test_integer_for_prime($p)) {
push @prime_factors, $p;
next;
}
#
my $d = pollard_rho_simple($p);
my $d = pollard_rho_strong($p);
if ($d == $p) {
push @factors, $d;
} else {
push @factors, $d;
push @factors, int($p / $d);
}
}
return \@prime_factors;
}
sub test_integer_for_prime {
my $p = shift;
my @probes = qw[ 2 3 5 7 11 13 17 ];
foreach my $a (@probes) {
return 1 if $a == $p;
}
my ($k, $q) = (0, $p - 1);
while (! ($q & 1)) {
$q >>= 1;
$k += 1;
}
my ($a_raised_to_q, $a_raised_to_jq, $primeflag);
foreach my $a (@probes) {
my ($base,$exponent) = ($a,$q);
my $a_raised_to_q = 1;
while ((int($exponent) > 0)) {
$a_raised_to_q = ($a_raised_to_q * $base) % $p
if int($exponent) & 1;
$exponent = $exponent >> 1;
$base = ($base * $base) % $p;
}
next if $a_raised_to_q == 1;
next if ($a_raised_to_q == ($p - 1)) && ($k > 0);
$a_raised_to_jq = $a_raised_to_q;
$primeflag = 0;
foreach my $j (0 .. $k - 2) {
$a_raised_to_jq = ($a_raised_to_jq ** 2) % $p;
if ($a_raised_to_jq == $p-1) {
$primeflag = 1;
last;
}
67
#(F1)
#(F2)
#(F3)
#(F4)
#(F5)
#(F6)
#(F8)
#(F9)
#(F10)
#(F11)
#(F12)
#(F13)
#(F14)
#(F15)
#(F16)
#(F17)
#(F18)
#(P1)
#(P2)
#(P3)
#(P4)
#(P5)
#(P6)
#(P7)
#(P8)
#(P9)
#(P10)
#(P11)
#(P12)
#(P13)
#(P14)
#(P15)
#(P16)
#(P17)
#(P18)
#(P19)
#(P20)
#(P21)
#(P22)
#(P23)
#(P24)
#(P25)
#(P26)
Computer and Network Security by Avi Kak
Lecture 12
}
return 0 if ! $primeflag;
#(P27)
}
my $probability_of_prime = 1 - 1.0/(4 ** scalar(@probes));
return $probability_of_prime;
#(P28)
#(P29)
}
sub pollard_rho_simple {
my $p = shift;
my @probes = qw[ 2 3 5 7 11 13 17 ];
foreach my $a (@probes) {
return $a if $p % $a == 0;
}
my $d = 1;
my $a = 2 + int(rand($p));
my @random_num = ($a);
while ($d == 1) {
my $b = 2 + int(rand($p));
foreach my $a (@random_num) {
$d = gcd($a - $b, $p);
last if $d > 1;
}
push @random_num, $b;
}
return $d;
}
#(Q1)
#(Q2)
#(Q3)
#(Q4)
#(Q5)
sub pollard_rho_strong {
my $p = shift;
my @probes = qw[ 2 3 5 7 11 13 17 ];
foreach my $a (@probes) {
return $a if $p % $a == 0;
}
my $d = 1;
my $a = 2 + int(rand($p));
my $c = 2 + int(rand($p));
my $b = $a;
while ($d == 1) {
$a = ($a * $a + $c) % $p;
$b = ($b * $b + $c) % $p;
$b = ($b * $b + $c) % $p;
$d = gcd($a - $b, $p);
last if $d > 1;
}
return $d;
}
#(R1)
#(R2)
#(R3)
#(R4)
sub gcd {
my ($a,$b) = @_;
while ($b) {
($a,$b) = ($b, $a % $b);
}
return $a;
}
#(Q6)
#(Q7)
#(Q8)
#(Q9)
#(Q10)
#(Q11)
#(Q12)
#(Q13)
#(Q14)
#(Q15)
#(R5)
#(R6)
#(R6)
#(R7)
#(R8)
#(R9)
#(R10)
#(R11)
#(R12)
#(R13)
#(R14)
#(S1)
#(S2)
#(S3)
#(S4)
#(S5)
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Computer and Network Security by Avi Kak
Lecture 12
– If you call the above script with the argument shown below
Factorize.pl 1844674407
the script will return the answer shown below:
Factors of 1844674407:
3 ^ 2
204963823 ^ 1
– On the other hand, if you call this script for a large integer,
as in
Factorize.pl 18446744073709551617
the script will come back with the error message:
Your number is too large for factorization by this script.
Instead, try the script ’FactorizeWithBigInt.pl’
This error message is triggered by the statement in line (A3)
of the script where we compare the user-supplied integer with
the largest integer that can be stored in 4 bytes.
– That brings me to a Math::BigInt variant of the Perl script
shown above in order to deal with arbitrarily large
integers. Although the Math::BigInt library is now a part of
the Perl core, it is somewhat awkward to use unlike what is the
case with Python where transitioning to the big-number representation happens under the hood. When using Math::BigInt,
all operations — addition, multiplication, exponentiation, modular multiplication, modular exponentiation, and so on — require calls to this module’s API.
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Computer and Network Security by Avi Kak
Lecture 12
– In the script that is shown below, we immediately convert
the user supplied integer as a command-line argument into
its Math::BigInt representation in line (A5). As stated in my
introduction to the Pollard-Rho algorithm, the algorithm requires randomly generated integers whose differences, if found
coprime to the integer that is being factorized, then become
the factors you are looking for. For the script Factorize.pl
shown above, we could call on Perl’s native rand() function to
supply us with those random numbers. [Since we upper-bounded
the integers to be factorized in that script to the largest that can be
stored in 4 bytes and since that is also the upper bound on the numbers that rand() can return, the behavior of rand() is consistent with
what the script Factorize.pl is capable of.]
However, when you
are dealing with arbitrarily large integers, you need a random
number generator commensurate with such numbers. That is
the reason for importing the Math::BigInt::Random::OO in line
(A2).
#!/usr/bin/env perl
##
##
##
FactorizeWithBigInt.pl
Author: Avi Kak
Date: February 21, 2016
##
##
Uncomment line (F13) and comment out line (F14) if you want to see the results
with the simpler form of the Pollard-Rho algorithm.
use
use
use
use
strict;
warnings;
Math::BigInt;
Math::BigInt::Random::OO;
########################### class FactorizeWithBigInt
package FactorizeWithBigInt;
sub new {
my ($class, $num) = @_;
##########################
#(A1)
#(A2)
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Computer and Network Security by Avi Kak
bless {
num =>
}, $class;
Lecture 12
#(A3)
#(A4)
#(A5)
int($num),
}
sub factorize {
my $self = shift;
my $n = $self->{num};
my @prime_factors = ();
my @factors;
push @factors, $n;
while (@factors > 0) {
my $p = pop @factors;
if ($self->test_integer_for_prime($p)) {
my $pnum = $p->numify();
push @prime_factors, $p;
next;
}
#
my $d = $self->pollard_rho_simple($p);
my $d = $self->pollard_rho_strong($p);
if ($d->copy()->bacmp($p->copy()) == 0) {
push @factors, $d;
} else {
push @factors, $d;
my $div = $p->copy()->bdiv($d->copy());
push @factors, $div;
}
}
return \@prime_factors;
}
sub test_integer_for_prime {
my $self = shift;
my $p = shift;
return 0 if $p->is_one();
my @probes = qw[ 2 3 5 7 11 13 17 ];
foreach my $a (@probes) {
$a = Math::BigInt->new("$a");
return 1 if $p->bcmp($a) == 0;
return 0 if $p->copy()->bmod($a)->is_zero();
}
my ($k, $q) = (0, $p->copy()->bdec());
while (! $q->copy()->band( Math::BigInt->new("1"))) {
$q->brsft( 1 );
$k += 1;
}
my ($a_raised_to_q, $a_raised_to_jq, $primeflag);
foreach my $a (@probes) {
my $abig = Math::BigInt->new("$a");
my $a_raised_to_q = $abig->bmodpow($q, $p);
next if $a_raised_to_q->is_one();
my $pdec = $p->copy()->bdec();
next if ($a_raised_to_q->bcmp($pdec) == 0) && ($k > 0);
$a_raised_to_jq = $a_raised_to_q;
$primeflag = 0;
71
#(F1)
#(F2)
#(F3)
#(F4)
#(F5)
#(F6)
#(F7)
#(F8)
#(F9)
#(F10)
#(F11)
#(F12)
#(F13)
#(F14)
#(F15)
#(F16)
#(F17)
#(F18)
#(F19)
#(F20)
#(F21)
#(P1)
#(P2)
#(P3)
#(P4)
#(P5)
#(P6)
#(P7)
#(P8)
#(P9)
#(P10)
#(P11)
#(P12)
#(P13)
#(P14)
#(P15)
#(P16)
#(P17)
#(P18)
#(P19)
#(P20)
#(P21)
#(P22)
Computer and Network Security by Avi Kak
Lecture 12
foreach my $j (0 .. $k - 2) {
my $two = Math::BigInt->new("2");
$a_raised_to_jq = $a_raised_to_jq->copy()->bmodpow($two, $p);
if ($a_raised_to_jq->bcmp( $p->copy()->bdec() ) == 0 ) {
$primeflag = 1;
last;
}
}
return 0 if ! $primeflag;
}
my $probability_of_prime = 1 - 1.0/(4 ** scalar(@probes));
return $probability_of_prime;
#(P23)
#(P24)
#(P25)
#(P26)
#(P27)
#(P28)
#(P29)
#(P30)
#(P31)
}
sub pollard_rho_simple {
my $self = shift;
my $p = shift;
my @probes = qw[ 2 3 5 7 11 13 17 ];
foreach my $a (@probes) {
my $abig = Math::BigInt->new("$a");
return $abig if $p->copy()->bmod($abig)->is_zero();
}
my $d = Math::BigInt->bone();
my $randgen = Math::BigInt::Random::OO->new( max => $p );
my $a = Math::BigInt->new();
unless ($a->numify() >= 2) {
$a = $randgen->generate(1);
}
my @random_num = ($a);
while ($d->is_one()) {
my $b = Math::BigInt->new();
unless ($b->numify() >= 2) {
$b = $randgen->generate(1);
}
foreach my $a (@random_num) {
$d = Math::BigInt::bgcd($a->copy()->bsub($b),$p);
last if $d->bacmp(Math::BigInt->bone()) > 0;
}
push @random_num, $b;
}
return $d;
}
sub pollard_rho_strong {
my $self = shift;
my $p = shift;
my @probes = qw[ 2 3 5 7 11 13 17 ];
foreach my $a (@probes) {
my $abig = Math::BigInt->new("$a");
return $abig if $p->copy()->bmod($abig)->is_zero();
}
my $d = Math::BigInt->bone();
my $randgen = Math::BigInt::Random::OO->new( max => $p );
my $a = Math::BigInt->new();
unless ($a->numify() >= 2) {
72
#(Q1)
#(Q2)
#(Q3)
#(Q4)
#(Q5)
#(Q6)
#(Q7)
#(Q8)
#(Q9)
#(Q10)
#(Q11)
#(Q12)
#(Q13)
#(Q14)
#(Q15)
#(Q16)
#(Q17)
#(Q18)
#(Q19)
#(Q20)
#(Q21)
#(Q22)
#(R1)
#(R2)
#(R3)
#(R4)
#(R5)
#(R6)
#(R7)
#(R8)
#(R9)
#(R10)
#(R11)
Computer and Network Security by Avi Kak
Lecture 12
$a = $randgen->generate(1);
}
$randgen = Math::BigInt::Random::OO->new( max => $p );
my $c = Math::BigInt->new();
unless ($c->numify() >= 2) {
$c = $randgen->generate(1);
}
my $b = $a->copy();
while ($d->is_one()) {
$a->bmuladd($a->copy(), $c->copy())->bmod($p);
$b->bmuladd($b->copy(), $c->copy())->bmod($p);
$b->bmuladd($b->copy(), $c->copy())->bmod($p);
$d = Math::BigInt::bgcd( $a->copy()->bsub($b), $p );
last if $d->bacmp(Math::BigInt->bone()) > 0;
}
return $d;
#(R12)
#(R13)
#(R14)
#(R15)
#(R16)
#(R17)
#(R18)
#(R19)
#(R20)
#(R21)
#(R22)
#(R23)
#(R24)
}
#################################
package main;
main
######################################
unless (@ARGV) {
1;
} else {
my $p = shift @ARGV;
$p = Math::BigInt->new( "$p" );
my $factorizer = FactorizeWithBigInt->new($p);
my @factors = @{$factorizer->factorize()};
my %how_many_of_each;
map {$how_many_of_each{$_}++} @factors;
print "\nFactors of $p:\n";
foreach my $factor (sort {$a <=> $b} keys %how_many_of_each) {
print "
$factor ^ $how_many_of_each{$factor}\n";
}
}
#(M1)
#(M2)
#(M3)
#(M2)
#(M3)
#(M4)
#(M5)
#(M6)
#(M7)
#(M8)
#(M9)
#(M10)
– To demonstrate the script shown above in action, if you call
FactorizeWithBigInt.pl
123456789123456789123456789123456789123456789123456789
the script returns the following factorization:
Factors of 123456789123456789123456789123456789123456789123456789:
3 ^ 3
7 ^ 1
11 ^ 1
13 ^ 1
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Computer and Network Security by Avi Kak
Lecture 12
19 ^ 1
757 ^ 1
3607 ^ 1
3803 ^ 1
52579 ^ 1
70541929 ^ 1
14175966169 ^ 1
440334654777631 ^ 1
– The Pollard-ρ algorithm is based on John Pollard’s article “A
Monte Carlo Method for Factorization,” BIT, pp. 331-334.
A more efficient variation on Pollard’s method was published
by Richard Brent: “An Improved Monte Carlo Factorization Algorithm,” in the same journal in 1980.
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Computer and Network Security by Avi Kak
Lecture 12
12.10: FACTORIZATION OF LARGE
NUMBERS: THE OLD RSA FACTORING
CHALLENGE
• Since the security of the RSA algorithm is so critically dependent
on the difficulty of finding the prime factors of a large number,
RSA Labs (http://www.rsasecurity.com/rsalabs/) used
to sponsor a challenge to factor the numbers supplied by them.
• The challenge generated a lot of excitement when it was active.
Many of the large numbers put forward by RSA Labs for factoring
have still not been factored and are not expected to be factored
any time soon.
• Given the historical importance of this challenge and the fact
that many of the numbers have not yet been factored makes it
interesting to review the state of the challenge today.
• The challenges are denoted
RSA-XXX
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Computer and Network Security by Avi Kak
Lecture 12
where XXX stands for the number of bits needed for a binary representation of the number to be factored in the round of
challenges starting with RSA − 576.
• Let’s look at the factorization of the number in the RSA-200
challenge (200 here refers to the number of decimal digits):
RSA-200 =
2799783391122132787082946763872260162107044678695
5428537560009929326128400107609345671052955360856
0618223519109513657886371059544820065767750985805
57613579098734950144178863178946295187237869221823983
Its two factors are
35324619344027701212726049781984643686711974001976250
23649303468776121253679423200058547956528088349
79258699544783330333470858414800596877379758573642
19960734330341455767872818152135381409304740185467
RSA-200 was factored on May 9, 2005 by Bahr, Boehm, Franke,
and Kleinjung of Bonn University and Max Planck Institute.
• Here is a description of RSA-576:
Name:
RSA-576
Prize:
$10000
Digits:
174
Digit Sum:
785
188198812920607963838697239461650439807163563379
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Computer and Network Security by Avi Kak
Lecture 12
417382700763356422988859715234665485319060606504
743045317388011303396716199692321205734031879550
656996221305168759307650257059
RSA-576 was factored on Dec 3, 2003 by using a combination of
lattice sieving and line sieving by a team of researchers (Franke,
Kleinjung, Montgomery, te Riele, Bahr, Leclair, Leyland, and
Wackerbarth) working at Bonn University, Max Planck Institute,
and some other places.
• Here is a description of RSA-640:
Name:
RSA-640
Prize:
$20000
Digits:
193
Digit Sum:
806
31074182404900437213507500358885679300373460228
42727545720161948823206440518081504556346829671
72328678243791627283803341547107310850191954852
90073377248227835257423864540146917366024776523
46609
RSA-640 was factored on November 2, 2005 by the same team
that solved RSA-576. Took over five months of calendar time.
• RSA-768, shown below, was factored in December 2009 by T.
Kleinjung, K. Aoki, J. Franke, A. Lenstra, E. Thome, J Bos, P.
Gaudry, A. Kruppa, P. Montgommery, D. Osvik, H. te Riele,
A. Timofeev, and P. Zimmerman. This is the largest modulus
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Computer and Network Security by Avi Kak
Lecture 12
that has been factored to date. This factorization resulted from
a multi-year effort in distributed computing.
Name:
RSA-768
Prize:
$50000
(retracted)
Digits:
232
Digit Sum:
1018
12301866845301177551304949583849627207728535695
95334792197322452151726400507263657518745202199
78646938995647494277406384592519255732630345373
15482685079170261221429134616704292143116022212
40479274737794080665351419597459856902143413
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Computer and Network Security by Avi Kak
Lecture 12
12.10.1: The Old RSA Factoring Challenge: Numbers
Not Yet Factored
Name:
RSA-896
Prize:
$75000
(retracted)
Digits:
270
Digit Sum:
1222
41202343698665954385553136533257594817981169984
43279828454556264338764455652484261980988704231
61841879261420247188869492560931776375033421130
98239748515094490910691026986103186270411488086
69705649029036536588674337317208131041051908642
54793282601391257624033946373269391
Name:
RSA-1024
Prize:
$100000
(retracted)
Digits:
309
Digit Sum:
1369
135066410865995223349603216278805969938881475605
667027524485143851526510604859533833940287150571
909441798207282164471551373680419703964191743046
496589274256239341020864383202110372958725762358
509643110564073501508187510676594629205563685529
475213500852879416377328533906109750544334999811
150056977236890927563
Name:
Prize:
RSA-1536
$150000
(retracted)
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Computer and Network Security by Avi Kak
Lecture 12
Digits:
463
Digit Sum:
2153
184769970321174147430683562020016440301854933866
341017147178577491065169671116124985933768430543
574458561606154457179405222971773252466096064694
607124962372044202226975675668737842756238950876
467844093328515749657884341508847552829818672645
133986336493190808467199043187438128336350279547
028265329780293491615581188104984490831954500984
839377522725705257859194499387007369575568843693
381277961308923039256969525326162082367649031603
6551371447913932347169566988069
Name:
RSA-2048
Prize:
$200000
(retracted)
Digits:
617
Digit Sum:
2738
2519590847565789349402718324004839857142928212620
4032027777137836043662020707595556264018525880784
4069182906412495150821892985591491761845028084891
2007284499268739280728777673597141834727026189637
5014971824691165077613379859095700097330459748808
4284017974291006424586918171951187461215151726546
3228221686998754918242243363725908514186546204357
6798423387184774447920739934236584823824281198163
8150106748104516603773060562016196762561338441436
0383390441495263443219011465754445417842402092461
6515723350778707749817125772467962926386356373289
9121548314381678998850404453640235273819513786365
64391212010397122822120720357
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Computer and Network Security by Avi Kak
Lecture 12
12.11: THE RSA ALGORITHM: SOME
OPERATIONAL DETAILS
• The main goal of this section is to explain how the public and the
private keys — which theoretically speaking are merely pairs of
integers [n, e] and [n, d], respectively, — are actually represented
in the memory of a computer. As you will see, the representation
used depends on the protocol. The key representation in the SSH
protocol is, for example, very different from the key representation in the TLS/SSL protocol. However, before getting to the
key representation issues, what follows are some very important
general comments about the RSA algorithm.
• The size of the key in the RSA algorithm typically refers to the
size of the modulus integer in bits. In that sense, the phrase
“key size” in the context of RSA is a bit of a misnomer. As
you now know, the actual keys in RSA are the public key [n, e]
and the private key [n, d]. In addition to depending on the size of
the modulus, the key sizes obviously depend on the values chosen
for e and d.
• Consider the case of an RSA implementation that provides 1024
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Computer and Network Security by Avi Kak
Lecture 12
bits of security. So we are talking about an implementation of
the RSA algorithm that uses a 1024 bit modulus. [It is interesting to
reflect on the fact that 1024 bits can be stored in only 128 bytes in the memory of a computer (and that
translates into a 256-character hex string if we had to print out the 128 bytes for visual display), yet
the decimal value of the integer represented by these 128 bytes can be monstrously large.
] Here
is an example of such a decimal number:
896648260163177445892450830685346881485335435
598887985722112773321881386436681238522440572
201181538908178518569358459456544005330977672
121582110702985339908050754212664722269478671
818708715560809784221316449003773512418972397
715186575579269079705255036377155404327546356
26323200716344058408361871194193919999
There are 359 decimal digits in this very large integer. [It is trivial
to generate arbitrarily large integers in Python since the language places no limits on
the size of the integer. I generated the above number by simply setting a variable to a
random 256 character hex string by a statement like
num = 0x7fafdbff7fe0f9ff7.... 256 hex characters ...... ff7fffda5f
and then just calling ’print num’.]
The above example should again
remind you of the exponential relationship between what it takes
to represent an integer in the memory of a computer and the
value of that integer.
• RSA Laboratories recommends that the two primes that compose
the modulus should be roughly of equal length. So if you want
to use 1024-bit RSA encryption, that means that your modulus
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Computer and Network Security by Avi Kak
Lecture 12
integer will have a 1024 bit presentation, and that further means
that you’d need to generate two primes that are roughly 512 bits
each.
• Doubling the size of the key will, in general, increase the time
required for public key operations (as needed for encryption or
signature verification) by a factor of four and increase the time
taken by private key operations (decryption and signing) by a factor of 8. Public key operations are not as affected as the private
key operations when you double the size of the key is because the
public key exponent e does not have to change as the key size increases. On the other hand, the private key exponent d changes in
direct proportion to the size of the modulus. The key generation
time goes up by a factor of 16 as the size of the key (meaning the
size of the modulus) is doubled. But key generation is a relatively
infrequent operation. (Ref.: http://www.rsa.com/rsalabs)
• The public and the private keys are stored in particular formats
specified by various protocols. For the public key, in addition to
storing the encryption exponent and the modulus, the key may
also include information such as the time period of validity,
the name of the algorithm used for key generation, etc. For
the private key, in addition to storing the decryption exponent
and the modulus, the key may include additional information
along the same lines as for the public key, and, additionally, the
corresponding public key also. Typically, the formats call for the
keys to be stored using Base64 encoding so that they can be
83
Computer and Network Security by Avi Kak
Lecture 12
displayed using printable characters. (See Lecture 2 on Base64
encoding.) To see such keys, you could, for example, experiment
with the following function:
ssh-keygen -t rsa
The public and the private keys returned by this call, when stored
appropriately, will allow your laptop to establish SSH connections
with machines elsewhere from virtually anywhere in the world
(unless a local firewall blocks SSH traffic) without you having
to log in explicitly with a password. [You can also replace ‘rsa’
with ‘dsa’ in the above call. The flag ‘dsa’ refers to the Digital Signature Algorithm that
typically uses the ElGamal protocol (see Section 13.6 of Lecture 13 for ElGamal) for
generating the key pairs. A call such as above will ask you for a passphrase, but you can
ignore it if you wish. The above call will store the private key in the file .ssh/id rsa of
the home account in your laptop. The public key will be deposited in a file that will be
named .ssh/id rsa.pub. Now all you have to do is to copy the public key into
the file .ssh/authorized keys of any of the remote machines to which you
want SSH access without the bother of having to log in with a password.]
• Here is an example of a private key in the .ssh/id_rsa file of a
now retired machine. Note that it is in Base64 encoding.
-----BEGIN RSA PRIVATE KEY----MIIEogIBAAKCAQEA5amriY96HQS8Y/nKc8zu3zOylvpOn3vzMmWwrtyDy+aBvns4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Computer and Network Security by Avi Kak
Lecture 12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-----END RSA PRIVATE KEY-----
• And here is an example of the public key that goes with the above
private key
ssh-rsa AAAAB3NzaC1yc2EAAAABIwAAAQEA5amriY96HQS8Y/nKc8zu3zOylvp
On3vzMmWwrtyDy+aBvns4UC1RXoaD9rDKqNNMCBAQwWDsYwCAFsrBzbxRQONHeP
X8lRWgM87MseWGlu6WPzWGiJMclTAO9CTknplG9wlNzLQBj3dP1M895iLF6jvJ7
GR+V3CRU6UUbMmRvgPcsfv6ec9RRPm/B8ftUuQICL0jt4tKdPG45PBJUylHs71F
uE9FJNp01hrj1EMFObNTcsy9zuis0YPyzArTYSOUsGglleExAQYi7iLh17pAa+y
6fZrGLsptgqryuftN9Q4NqPuTiFjlqRowCDU7sSxKDgU7bzhshyVx3+pzXO4D2Q
== kak@pixie
• In general, the format used for storing a key is specific to each
protocol. The public key shown above is for the SSH protocol
as described in RFC 4253. An SSH public key stores the following three fields separated by white space: (1) the key type;
(2) a chunk of Base64 encoded data; and (3) A comment. In
the public key that I showed above, the first and the third fields
are, respectively, the strings ‘ssh-rsa’ and ‘kak@pixie’. What is
in-between the two is the Base64 encoded data that holds the
85
Computer and Network Security by Avi Kak
Lecture 12
public exponent and the modulus integers. After you Base64decode this string, you end up with a stream of bytes for three
<length data> records. These three records hold the following
three pieces of information: (1) Algorithm name (which would
be the same as the key-type you would have seen in the first field
of the public key; (2) the RSA public exponent; and (3) the RSA
modulus. In each record, the length value is stored in the first
four bytes in the Big-endian form. [Therefore, in order to extract the (e, n) integers from the key shown above, we must scan the byte stream that we get after Base64 decoding of the middle
field shown above. We look at the first four bytes to see how many subsequent bytes hold the name of the algorithm. After we have read off those bytes, we again look at the next four bytes to find out how many subsequent
] Shown below is a
Python script that extracts the public exponent and the modulus
stored in an SSH RSA public key. In line with the note in blue,
the script first separates the three field in the key by splitting it
on white space. It then applies Base64 decoding to the middle
field since that’s where the public exponent and the modulus are
stored. Subsequently, it scans the stream of decoded bytes for the
<length,value> records under the assumption that the length
of the value is always placed in the first four bytes of each record.
bytes hold the public exponent; and so on for extracting the modulus integer.
#!/usr/bin/env python
##
##
##
extract_sshpubkey_params.py
Author: Avi Kak
Date:
February 11, 2013
import sys
import base64
import BitVector
if len(sys.argv) != 2:
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Computer and Network Security by Avi Kak
Lecture 12
sys.stderr.write("Usage: %s
<public key file>\n" % sys.argv[0])
sys.exit(1)
keydata = base64.b64decode(open(sys.argv[1]).read().split(None)[1])
bv = BitVector.BitVector( rawbytes = keydata )
parts = []
while bv.length() > 0:
bv_length = int(bv[:32])
# read 4 bytes for length of data
data_bv = bv[32:32+bv_length*8]
# read the data
parts.append(data_bv)
bv.shift_left(32+bv_length*8)
# shift the starting BV and
bv = bv[0:-32-bv_length*8]
#
and truncate its length
public_exponent = int(parts[1])
modulus = int(parts[2])
print "public exponent: ", public_exponent
print "modulus: ", modulus
• If I invoke the above script on my public SSH RSA key in
~/.ssh/id_rsa.pub by
extract_sshpubkey_params.py ~/.ssh/id_rsa.pub
I get the following output:
public exponent:
modulus:
35
28992239265965680130833686108835390387986295644147105350109222053494471862488069515097328563379
83891022841669525585184878497657164390613162380624769814604174911672498450880421371197440983388
47257142771415372626026723527808024668042801683207069068148652181723508612356368518824921733281
43920627731421841448660007107587358412377023141585968920645470981284870961025863780564707807073
26000355974893593324676938927020360090167303189496460600023756410428250646775191158351910891625
48335568714591065003819759709855208965198762621002125196213207135126179267804883812905682728422
31250173298006999624238138047631459357691872217
• The SSL/TLS public and private keys, as also the SSH RSA
private keys, are, on the other hand, stored using a more elaborate
procedure: The key information is first encoded using Abstract
Syntax Notation (ASN) according to the ASN.1 standard and the
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Computer and Network Security by Avi Kak
Lecture 12
resulting data structure DER-encoded into a byte stream. (DER
standards for ‘Distinguished Encoding Rules’ — it’s a part of
the ASN.1 standard.) Finally, the byte stream thus generated is
turned into a printable representation by Base64 encoding. [The
ASN.1 standard, along with one of its transfer encodings such as DER, accomplishes the same thing for complex
data structures in a binary format that the XML standard does in a textual format. You can certainly convert
XML representations into binary formats, but the resulting encoding will, in general, be much longer than
those produced by ASN.1. Let’s say you wish to represent all of your assets in a manner that would be directly
readable by different computing platforms and different programming languages. A record of your assets is
likely to consist of the names of the financial institutions and the value of the assets held by them, a listing of
your fixed assets, such as real estate properties and their worth, etc. In general, such data will require a tree
representation in which the various nodes may stand for the names of the financial institutions or the names
of the assets and the children of the leaf nodes would consist of asset values. The values for some of the nodes
may be in the form of ordered lists, unordered lists (sets), key-value pairs, etc. ASN.1 creates compact byte
level representations for such structures that is portable across platforms and languages. Just to give you a
small taste of the flexibility of ASN.1 representation, it places no constraints on the size of any of the symbolic
entities or any of the numerical values. And to also give you a taste of the secret to the sauce, when ASN.1 is
used with BER (Basic Encoding Rules) encoding, each node of the tree is represented by three blocks of bytes:
(1) Identification block of an unlimited number of bytes; (2) Length block of an unlimited number of bytes; and
(3) Value block of an unlimited number of bytes.The important thing to note here is there are no constraints
on how many bytes are taken up by each of the thee blocks. How does ASN.1 accomplish that? It’s all done
by using high-end bytes to carry information about bytes further downstream. For example, if the length is to
be represented by a single byte, then the value of length must not exceed 128. However, if the value of length
is 128 or greater, then the most significant bit of the first byte must be set to 1 and the trailing bits must tell
us how many of the following bytes are being used for storing the length information. Similar rules are used
]
for the other blocks to permit them to be of arbitrary length.
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Lecture 12
• To generate the private and public keys for the SSL/TLS protocol
you can use the OpenSSL library in the following manner:
openssl genrsa -out myprivate.pem 1024
openssl rsa -in myprivate.pem -pubout > mypublic.pem
where the first command creates a private key for a 1024 bit
modulus and the second then gives you the corresponding public
key. The private key will be deposited in the file myprivate.pem
and the public key in the file mypublic.pem.
• If you want to see the modulus and the public exponent used in
the public key, you can execute
openssl rsa -pubin -inform PEM -text -noout < mypublic.pem
• As mentioned earlier, SSL/TLS keys are stored (and transmitted)
by first encoding them with the abstract notation of ASN.1, turning the resulting structure into a byte stream with DER encoding,
and, finally, making this byte stream printable with Base64 encoding. [The standards documents that address the formatting of such keys are RFC 3447 and 4716.]
The ASN.1 representation of a public RSA key is given by:
SEQUENCE {
SEQUENCE {
OBJECT IDENTIFIER rsaEncryption (1 2 840 113549 1 1 1),
NULL
}
BIT STRING {
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Computer and Network Security by Avi Kak
Lecture 12
RSAPublicKey ::= SEQUENCE {
modulus
INTEGER,
publicExponent
INTEGER
}
-- n
-- e
}
}
where ”1 2 840 113549 1 1 1” is the ANS.1 specified object ID
for rsaEncryption and where n is the modulus and e the public exponent. [In symbolic depictions of ASN.1 data structures,
what comes after a double hyphen is a comment.] The terms
SEQUENCE, BIT STRING, etc. are some of the ASN.1 keywords.
[Note that the ASN.1 keywords, such as ‘SEQUENCE’, ‘BIT STRING’, etc., that you see in the data structure
above determine how the data bearing bytes are laid out in the byte-stream representation of the object. These
keywords themselves do not appear directly in their symbolic forms in the byte level representation of a key.
An agent receiving such a key would know its ”schema” from the object identifier and would thus be able to
]
decode the bytes.
• The ASN.1 representation for a private key is given by (where
we have suppressed ancillary information related to the object
identity, etc.):
RSAPrivateKey ::= SEQUENCE {
version
Version,
modulus
INTEGER,
publicExponent
INTEGER,
privateExponent INTEGER,
prime1
INTEGER,
prime2
INTEGER,
exponent1
INTEGER,
exponent2
INTEGER,
90
--------
n
e
d
p
q
d mod (p-1)
d mod (q-1)
Computer and Network Security by Avi Kak
coefficient
otherPrimeInfos
Lecture 12
INTEGER,
-- (inverse of q) mod p
OtherPrimeInfos OPTIONAL
}
where n is the modulus, e the public exponent, d the private
exponent, p and q the two primes whose product is the modulus.
The rest of the fields are used in the modular exponentiation that
is carried out for decryption.
• In Perl, you can use the Convert::ASN1 module for creating
an ASN.1 encoded representation of a data structure and for its
transformation into a byte stream with BER or DER encodings.
In Python, you can do the same with the pyasn1 library.
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Lecture 12
12.12: IN SUMMARY . . .
• Assuming that you are using the best possible random number
generators to create candidates for the primes that are needed
and that you also use a recent version of the RSA scheme that
is resistant to the chosen ciphertext attacks, the security of RSA
encryption depends critically on the difficulty of factoring large
integers.
• As integer factorization algorithms have become more and more
powerful over the years, RSA cryptography has had to rely on
increasingly larger values for the integer modulus and, therefore,
increasingly longer encryption keys.
• These days you are unlikely to use a key whose length is — or, to speak
more precisely, a modulus whose size is — shorter than 1024 bits for RSA. Some
people recommend 2048 or even 4096 bit keys. The following
table vividly illustrates how the key sizes compare for symmetrickey cryptography and RSA-based public-key cryptography for the
same level of cryptographic security [Values taken from NIST Special Publication
800-57, Recommendations for Key Management — Part 1,” by Elaine Barker et al.]
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Computer and Network Security by Avi Kak
Lecture 12
Symmetric Key Algorithm
Key Size for the
Symmetric Key Algorithm
Comparable RSA Key Length
for the Same Level of Security
2-Key 3DES
3-Key 3DES
112
168
1024
2048
AES-128
AES-192
AES-256
128
192
256
3072
7680
15360
• As you’d expect, the computational overhead of RSA encryption/decryption goes up as the size of the modulus integer increases.
• This makes RSA inappropriate for encryption/decryption of actual message content for high data-rate communication links.
• However, RSA is ideal for the exchange of secret keys that can
subsequently be used for the more traditional (and much faster)
symmetric-key encryption and decryption of the message content.
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Computer and Network Security by Avi Kak
Lecture 12
12.13: HOMEWORK PROBLEMS
1. The necessary condition for the encryption key e is that it be
coprime to the totient of the modulus. But, in practice, what is
e typically set to and why? (Obviously, now the burden falls on
ensuring the selected primes p and q are such that the necessary
condition on e is still satisfied.)
2. On the basis of the material presented in Sections 12.6, 12.7, 12.8
and 12.9, make your own assessment of the security vulnerabilities
of RSA that are important today, that could become important
in the next decade, and that could be important over the very
long term.
3. From the public key, we know the modulus n and the encryption
integer e. If a bad guy could figure out the totient of the modulus,
would that amount to breaking the code?
4. Following the steps outlined in Section 12.4, create an RSA block
cipher with 16 bits of encryption (implying that you will use a
16-bit number for the modulus n in your cipher). Do NOT use
the same primes for p and q that I used in my example in Section
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Computer and Network Security by Avi Kak
Lecture 12
12.4. Use the n and e part of the cipher for block encryption of
the 6-byte word “purdue”. Print out the encrypted word as a
12-character hex string. Next use the n and d part of the cipher
to decrypt the encrypted string.
5. Assume for the sake of argument that your RSA scheme is as
simple as the one outlined in the toy example of Section 12.4.
How do you think it is possible for an attacker to figure out the
message bytes from the ciphertext bytes without access to the
private exponent?
6. As you now know, in the RSA algorithm a message M is encrypted by calculating:
C
=
M e (mod n)
where n is the modulus.
Assume that you are using a 1024-bit RSA algorithm (meaning
that the modulus is of size 1024 bits) for encrypting your messages. Now let’s say that your enemy knows that your business
partners are in the habit of communicating with you with very
short messages — messages that involve very small values of M
compared to the size of the n = p × q modulus.
Since the enemy will know your public key, he will know that
what your business partner has sent you is C = M e where e is the
public exponent that the enemy would know about. Assuming
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Computer and Network Security by Avi Kak
Lecture 12
for the sake of convenience that e = 3, why can’t the enemy
decrypt the confidential message intended for you by just taking
the cube-root of C?
7. Programming Assignment:
To better understand the point made in Section 12.3.2 that a
small value, such as 3, for the encryption integer e is cryptographically unsafe, assume that a party A has sent the same
message M = 10 to three different recipients using the following
three public keys:
[29, 3]
[37, 3]
[41, 3]
In each public key, the first integer is the modulus n and the
second the encryption integer e. Now use the Chinese Remainder
Theorem of Section 11.7 in Lecture 11 to show how you can
reconstruct M 3, which in this case would be 1000, from the three
ciphertext values corresponding to the three public keys. [HINT:
If you are using Python, the ciphertext value in each case is returned by the built-in 3-argument function
pow(). For example, pow(M, 3, 29) will return the ciphertext integer C1 for the first public key shown above.
For each public key, we have Ci = M 3 mod ni where the three moduli are denoted n1 = 29, n2 = 37, and
n3 = 41. Now to solve the problem, you can reason as follows: Since n1 , n2 , and n3 are pairwise co-prime,
CRT allows us to reconstruct M 3 modulo N = n1 × n2 × n3 . This will require that you find Ni = N/ni for
i = 1, 2, 3. And then you would need to find the multiplicative inverse of each Ni modulo its corresponding ni .
Let Niinv denote this multiplicative inverse. You can use the Python multiplicative-inverse calculator shown
in Section 5.7 of Lecture 5 to calculate the Niinv values. Then, by CRT, you should be able to recover M 3 by
]
(C1 × N1 × N1inv + C2 × N2 × N2inv + C3 × N3 × N3inv ) mod N .
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Computer and Network Security by Avi Kak
Lecture 12
8. Programming Assignment:
Using the Python or the Perl version of the PrimeGenerator
class shown below and the multiplicative-inverse finding scripts
presented earlier in Section 5.7 of Lecture 5, write a script that
would constitute a “complete” implementation of a 64-bit RSA
algorithm. (As you now know from Section 12.7, a truly complete
implementation of RSA involves serious security considerations
related to padding, etc., that are beyond the scope of a homework
assignment. All you are being asked to do in this homework is to
address the basic mathematics of RSA.)
#!/usr/bin/env python
##
##
##
##
PrimeGenerator.py
Author: Avi Kak
Date: February 18, 2011
Modified Date: February 28, 2016
##
##
##
##
##
##
##
##
##
##
##
##
##
##
##
##
##
Call syntax:
PrimeGenerator.py
width_desired_for_bit_field_for_prime
For example, if you call
PrimeGenerator.py 32
you may get a prime that looks like 3262037833.
call
On the other hand, if you
PrimeGenerator.py 128
you may get a prime that looks like 338816507393364952656338247029475569761
IMPORTANT:
The two most significant are explicitly set for the prime that is
returned.
import sys
import random
############################
class PrimeGenerator
97
##############################
Computer and Network Security by Avi Kak
Lecture 12
class PrimeGenerator( object ):
#(A1)
def __init__( self, **kwargs ):
bits = debug = None
if ’bits’ in kwargs :
bits = kwargs.pop(’bits’)
if ’debug’ in kwargs :
debug = kwargs.pop(’debug’)
self.bits
=
bits
self.debug
=
debug
self._largest
=
(1 << bits) - 1
#(A2)
#(A3)
#(A4)
#(A5)
#(A6)
#(A7)
#(A8)
def set_initial_candidate(self):
candidate = random.getrandbits( self.bits )
if candidate & 1 == 0: candidate += 1
candidate |= (1 << self.bits-1)
candidate |= (2 << self.bits-3)
self.candidate = candidate
#(B1)
#(B2)
#(B3)
#(B4)
#(B5)
#(B6)
def set_probes(self):
self.probes = [2,3,5,7,11,13,17]
#(C1)
#(C2)
# This is the same primality testing function as shown earlier
# in Section 11.5.6 of Lecture 11:
def test_candidate_for_prime(self):
#(D1)
’returns the probability if candidate is prime with high probability’
p = self.candidate
#(D2)
if p == 1: return 0
#(D3)
if p in self.probes:
#(D4)
self.probability_of_prime = 1
#(D5)
return 1
#(D6)
if any([p % a == 0 for a in self.probes]): return 0
#(D7)
k, q = 0, self.candidate-1
#(D8)
while not q&1:
#(D9)
q >>= 1
#(D10)
k += 1
#(D11)
if self.debug: print("q = %d k = %d" % (q,k))
#(D12)
for a in self.probes:
#(D13)
a_raised_to_q = pow(a, q, p)
#(D14)
if a_raised_to_q == 1 or a_raised_to_q == p-1: continue
#(D15)
a_raised_to_jq = a_raised_to_q
#(D16)
primeflag = 0
#(D17)
for j in range(k-1):
#(D18)
a_raised_to_jq = pow(a_raised_to_jq, 2, p)
#(D19)
if a_raised_to_jq == p-1:
#(D20)
primeflag = 1
#(D21)
break
#(D22)
if not primeflag: return 0
#(D23)
self.probability_of_prime = 1 - 1.0/(4 ** len(self.probes))
#(D24)
return self.probability_of_prime
#(D25)
def findPrime(self):
self.set_initial_candidate()
if self.debug: print("
candidate is: %d" % self.candidate)
self.set_probes()
if self.debug: print("
The probes are: %s" % str(self.probes))
max_reached = 0
98
#(E1)
#(E2)
#(E3)
#(E4)
#(E5)
#(E6)
Computer and Network Security by Avi Kak
Lecture 12
while 1:
if self.test_candidate_for_prime():
if self.debug:
print("Prime number: %d with probability %f\n" %
(self.candidate, self.probability_of_prime) )
break
else:
if max_reached:
self.candidate -= 2
elif self.candidate >= self._largest - 2:
max_reached = 1
self.candidate -= 2
else:
self.candidate += 2
if self.debug:
print("
candidate is: %d" % self.candidate)
return self.candidate
####################################
if __name__ == ’__main__’:
main
#(E7)
#(E8)
#(E9)
#(E10)
#(E11)
#(E12)
#(E13)
#(E14)
#(E15)
#(E16)
#(E17)
#(E18)
#(E19)
#(E20)
#(E21)
#(E22)
######################################
if len( sys.argv ) != 2:
sys.exit( "Call syntax: PrimeGenerator.py width_of_bit_field" )
num_of_bits_desired = int(sys.argv[1])
generator = PrimeGenerator( bits = num_of_bits_desired )
prime = generator.findPrime()
print("Prime returned: %d" % prime)
#(M1)
#(M2)
#(M3)
#(M4)
#(M5)
#(M6)
If you make the following call to this script:
PrimeGenerator.py 64
the script will return a full-width 64-bit prime that will look like:
Prime returned: 17828589080991197309
On the other hand, a call like
PrimeGenerator.py 128
will return something like:
Prime returned: 290410362853346697538147183843312052911
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Computer and Network Security by Avi Kak
Lecture 12
For those of you will be doing this homework in Perl, here is a
Perl version of the above script:
#!/usr/bin/env perl
##
##
##
PrimeGenerator.pl
Author: Avi Kak
Date: February 26, 2016
##
##
##
##
##
##
##
##
##
##
##
##
##
##
##
##
##
Call syntax:
PrimeGenerator.pl
width_desired_for_bit_field_for_prime
For example, if you call
PrimeGenerator.pl 32
you may get a prime that looks like 3340094299.
call
On the other hand, if you
PrimeGenerator.pl 128
you may get a prime that looks like 333618953930748159614512936853740718827
IMPORTANT:
The two most significant are explicitly set for the prime that is
returned.
use strict;
use warnings;
use Math::BigInt;
############################
package PrimeGenerator;
class PrimeGenerator
sub new {
my ($class, %args) = @_;
bless {
_bits => int($args{bits}),
_debug => $args{debug} || 0,
_largest => (1 << int($args{bits})) - 1,
}, $class;
}
##############################
#(A1)
#(A2)
#(A3)
#(A4)
#(A5)
#(A6)
#(A7)
sub set_initial_candidate {
#(B1)
my $self = shift;
#(B2)
my @arr = map {my $x = rand(1); $x > 0.5 ? 1 : 0 } 0 .. $self->{_bits}-4;#(B3)
push @arr, 1;
#(B4)
unshift @arr, (1,1);
#(B6)
my $bstr = join ’’, split /\s/, "@arr";
#(B7)
#
$self->{candidate} = oct("0b".$bstr);
#(B8)
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Computer and Network Security by Avi Kak
Lecture 12
$self->{candidate} = Math::BigInt->from_bin($bstr);
#(B8)
}
sub set_probes {
my $self = shift;
$self->{probes} = [2,3,5,7,11,13,17];
}
#(C1)
#(C2)
#(C3)
# This is the same primality testing function as shown earlier
# in Section 11.5.6 of Lecture 11:
sub test_candidate_for_prime_with_bigint {
my $self = shift;
my $p = $self->{candidate};
return 0 if $p->is_one();
my @probes = @{$self->{probes}};
foreach my $a (@probes) {
$a = Math::BigInt->new("$a");
return 1 if $p->bcmp($a) == 0;
return 0 if $p->copy()->bmod($a)->is_zero();
}
my ($k, $q) = (0, $p->copy()->bdec());
while (! $q->copy()->band( Math::BigInt->new("1"))) {
$q->brsft( 1 );
$k += 1;
}
my ($a_raised_to_q, $a_raised_to_jq, $primeflag);
foreach my $a (@probes) {
my $abig = Math::BigInt->new("$a");
my $a_raised_to_q = $abig->bmodpow($q, $p);
next if $a_raised_to_q->is_one();
my $pdec = $p->copy()->bdec();
next if ($a_raised_to_q->bcmp($pdec) == 0) && ($k > 0);
$a_raised_to_jq = $a_raised_to_q;
$primeflag = 0;
foreach my $j (0 .. $k - 2) {
my $two = Math::BigInt->new("2");
$a_raised_to_jq = $a_raised_to_jq->copy()->bmodpow($two, $p);
if ($a_raised_to_jq->bcmp( $p->copy()->bdec() ) == 0 ) {
$primeflag = 1;
last;
}
}
return 0 if ! $primeflag;
}
my $probability_of_prime = 1 - 1.0/(4 ** scalar(@probes));
$self->{probability_of_prime} = $probability_of_prime;
return $probability_of_prime;
}
sub findPrime {
my $self = shift;
$self->set_initial_candidate();
print "
candidate is: $self->{candidate}\n" if $self->{_debug};
$self->set_probes();
print "
The probes are: @{$self->{probes}}\n" if $self->{_debug};
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#(D1)
#(D2)
#(D3)
#(D4)
#(D5)
#(D6)
#(D7)
#(D8)
#(D9)
#(D10)
#(D11)
#(D12)
#(D13)
#(D14)
#(D15)
#(D16)
#(D17)
#(D18)
#(D19)
#(D20)
#(D21)
#(D22)
#(D23)
#(D24)
#(D25)
#(D26)
#(D27)
#(D28)
#(D29)
#(D30)
#(D31)
#(E1)
#(E2)
#(E3)
#(E4)
#(E5)
#(E6)
Computer and Network Security by Avi Kak
Lecture 12
my $max_reached = 0;
while (1) {
if ($self->test_candidate_for_prime_with_bigint()) {
print "Prime number: $self->{candidate} with probability: " .
"$self->{probability_of_prime}\n" if $self->{debug};
last;
} else {
if ($max_reached ) {
$self->{candidate} -= 2;
} elsif ($self->{candidate} >= $self->{_largest} - 2) {
$max_reached = 1;
$self->{candidate} -= 2;
} else {
$self->{candidate} += 2;
}
}
}
return $self->{candidate};
#(E7)
#(E8)
#(E9)
#(E10)
#(E11)
#(E12)
#(E13)
#(E14)
#(E15)
#(E16)
#(E17)
#(E18)
#(E19)
#(E20)
}
1;
################################
package main;
main
########################################
unless (@ARGV) {
1;
} else {
my $bitfield_width = shift @ARGV;
my $generator = PrimeGenerator->new(bits => $bitfield_width);
my $prime = $generator->findPrime();
print "Prime returned: $prime\n";
}
#(M1)
#(M2)
#(M3)
#(M4)
#(M5)
#(M6)
#(M7)
A call such as the one shown below for generating a 256 bit prime
PrimeGenerator.pl 256
comes back with
Prime returned: 110683214729271322144990809842795090895043970651486233118696734813266440218909
9. Programming Assignment:
This assignment is also about implementing the RSA algorithm,
but now you are allowed to use modules from open-source libraries
102
Computer and Network Security by Avi Kak
Lecture 12
for some of the work. Because these libraries sit on top of highly
efficient C code, you should be able to test your implementation
for much larger moduli than what you used in the previous programming assignment. Write Perl or Python scripts that implement the RSA encryption and decryption algorithms. Do NOT
use the key-generator functions implemented in the modules of
the Perl/Python toolkits to find d for a given e. On the other
hand, you must use either the Python implementation shown in
Section 5.7 of Lecture 5 or your own implementation of the Extended Euclidean Algorithm to find the multiplicative inverses
you need. Feel free to use any other modules in the toolkits listed
below, or, for that matter, any other modules of you choice. However, you must list the modules used and where you found them
in the reference section of your code.
Python Cryptography Toolkit: http://www.amk.ca/python/code/crypto
Perl Crypt-RSA Toolkit: http://search.cpan.org/~ vipul/Crypt-RSA-1.57/lib/
Crypt/RSA.pm
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